mathplayer.weebly.commathplayer.weebly.com/uploads/3/8/0/0/38006537/ccgps_advanced_alg… · v i...

1 2 3 4 5 6 7 8 9 10

ISBN 978-0-8251-73

Copyright © 2014

J. Weston Walch, Publisher

Portland, ME 04103

www.walch.com

Printed in the United States of America

EDUCATIONWALCH

These materials may not be reproduced for any purpose.The reproduction of any part for an entire school or school system is strictly prohibited.

No part of this publication may be transmitted, stored, or recorded in any formwithout written permission from the publisher.

iiiTable of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Unit 3: Rational and Radical RelationshipsLesson 1: Operating with Rational Expressions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . U3-1Lesson 2: Solving Rational and Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . U3-37Lesson 3: Graphing Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U3-84Lesson 4: Graphing Radical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U3-146Lesson 5: Comparing Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . U3-174

Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . AK-1

Table of Contents

vIntroduction

Welcome to the CCGPS Advanced Algebra Student Resource Book. This book will help you learn how to use algebra, geometry, data analysis, and probability to solve problems. Each lesson builds on what you have already learned. As you participate in classroom activities and use this book, you will master important concepts that will help to prepare you for the EOCT and for other mathematics assessments and courses.

This book is your resource as you work your way through the Advanced Algebra course. It includes explanations of the concepts you will learn in class; math vocabulary and definitions; formulas and rules; and exercises so you can practice the math you are learning. Most of your assignments will come from your teacher, but this book will allow you to review what was covered in class, including terms, formulas, and procedures.

• In Unit 1: Inferences and Conclusions from Data, you will learn about summarizing and interpreting data and using the normal curve. You will explore populations, random samples, and sampling methods, as well as surveys, experiments, and observational studies. Finally, you will compare treatments and read reports.

• In Unit 2: Polynomial Functions, you will begin by exploring polynomial structures and operations with polynomials. Then you will go on to prove identities, graph polynomial functions, solve systems of equations with polynomials, and work with geometric series.

• In Unit 3: Rational and Radical Relationships, you will be introduced to operating with rational expressions. Then you will learn about solving rational and radical equations and graphing rational functions. You will solve and graph radical functions. Finally, you will compare properties of functions.

• In Unit 4: Exponential and Logarithmic Functions, you will start working with exponential functions and begin exploring logarithmic functions. Then you will solve exponential equations using logarithms.

• In Unit 5: Trigonometric Functions, you will begin by exploring radians and the unit circle. You will graph trigonometric functions, including sine and cosine functions, and use them to model periodic phenomena. Finally, you will learn about the Pythagorean Identity.

Introduction

Introductionvi

• In Unit 6: Mathematical Modeling, you will use mathematics to model equations and piecewise, step, and absolute value functions. Then, you will explore constraint equations and inequalities. You will go on to model transformations of graphs and compare properties within and between functions. You will model operating on functions and the inverses of functions. Finally, you will learn about geometric modeling.

Each lesson is made up of short sections that explain important concepts, including some completed examples. Each of these sections is followed by a few problems to help you practice what you have learned. The “Words to Know” section at the beginning of each lesson includes important terms introduced in that lesson.

As you move through your Advanced Algebra course, you will become a more confident and skilled mathematician. We hope this book will serve as a useful resource as you learn.

Lesson 1: Operating with Rational Expressions

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPS

U3-1Lesson 1: Operating with Rational Expressions

Common Core Georgia Performance Standards

MCC9–12.A.APR.6

MCC9–12.A.APR.7 (+)

Essential Questions

1. What does a rational expression look like?

2. How can a rational expression be rewritten into an equivalent form?

3. How is it possible to find the sum or difference of rational expressions?

4. How is it possible to find the product of rational expressions?

5. How is it possible to find the quotient of rational expressions?

WORDS TO KNOW

common denominator a quantity that is a shared multiple of the denominators of two or more fractions

least common denominator (LCD)

the least common multiple of the denominators of two or more fractions

rational expression an expression made of the ratio of two polynomials, in which a variable appears in the denominator

reciprocal a number that, when multiplied by the original number, has a product of 1

U3-2Unit 3: Rational and Radical Relationships

Recommended Resources

• Math.com. “Quadratic Equation Solver.”

http://www.walch.com/rr/00198

After entering the coefficients of a quadratic equation into this straightforward site, users can press the “Solve” button to generate appropriate roots. Detailed factors are not provided, and users must use the zeros to determine their own factors.

• Mathway. “Math Problem Solver.”


This resource is an online symbolic algebraic manipulator, capable of simplifying expressions. It can be a useful tool for tech-savvy individuals.

• WolframAlpha. “Simplify Rational Expressions.”


A powerful online symbolic manipulator, this resource can simplify complex rational expressions. It requires precise data entry to achieve valuable results.


Lesson 3.1.1: Adding and Subtracting Rational ExpressionsIntroduction

Expressions come in a variety of types, including rational expressions. A rational

expression is a ratio of two polynomials, in which a variable appears in the

denominator; for example, x

x

3

1+ is a rational expression. Working with rational

expressions can often be made easier by analyzing them to uncover more familiar (and

sometimes less complex) structures within them. In this section, you will explore some

of those structures.

Where one rational expression exists, another may as well. Extracting meaning from the context may require these expressions to be combined in order to determine a sum or difference. Combining rational expressions through addition or subtraction is not complex, though it does demand attention to detail. Rewriting equivalent fractions can often simplify the solution process.

Key Concepts

• Before adding or subtracting rational expressions, you must find a common denominator. A common denominator is a quantity that is a shared multiple of the denominators of two or more fractions.

• A common denominator can be determined by finding the product of

the denominators. For example, a common denominator of the rational

expression x

x

x

x x

3

1

2 1

22−+

++ −

is found by multiplying the two denominators,

x – 1 and x2 + x – 2: (x – 1)(x2 + x – 2) = x3 – 3x + 2.

• Once a common denominator has been found, it can be used to write equivalent rational expressions for each term of the sum (or difference, if subtracting).

• Using common denominators, the same rational expression as before, x

x

x

x x

3

1

2 1

22−+

++ −

, can be rewritten in an equivalent form:

x

x

x

x x

x

x

x x

x x

x

x x

x

x

3

1

2 1

2

3

1•

2

2

2 1

2•

1

12

2

2 2−+

++ −

=−

+ −+ −

++

+ −−−

x x x

x x

x x

x x

3 3 6

3 2

2 1

3 2

3 2

3

2

3=+ −− +

+− −− +


• Expressed with a common denominator, the sum of rational expressions is the

sum of the numerators: x

z

y

z

x y

z+ =

+. Thus, we can rewrite the expressions

over a single denominator, as shown.

x x x

x x

x x

x x

x x x x x

x x

3 3 6

3 2

2 1

3 2

3 3 6 2 1

3 2

3 2

3

2

3

3 2 2

3

+ −− +

+− −− +

=+ − + − −

− +

• Expressed with a common denominator, the difference of rational expressions

is the difference of the numerators: x

z

y

z

x y

z− =

−.

• The least common denominator (LCD) is the least common multiple of

the denominators of two or more fractions. In other words, it’s the smallest

possible common denominator. The LCD can be determined by finding the

product of all the unique factors of the denominator. Our sample rational

expression, x

x

x

x x

3

1

2 1

22−+

++ −

, is equivalent to the rational expression

x

x

x

x x

3

1

2 1

( 1)( 2)−+

+− +

. The LCD is (x – 1)(x + 2), or x2 – x – 2.

• Recall that a rational expression cannot include a value in the denominator that causes it to equal 0, since 0 is undefined in the denominator.


Guided Practice 3.1.1Example 1

Simplify the rational expression x

3

5

2+ .

1. Identify any invalid values of the expression.

The first term, 3

5, is composed entirely of constants; there are no

invalid values. With an x in the denominator of the second term, x

2,

we have to ensure that x ≠ 0 to keep this expression valid.

2. Find a common denominator.

The two denominators are 5 and x. Their product, 5x, is a useful common denominator.

3. Rewrite each term of the expression using the new denominator.

If the fraction’s denominator is not the common denominator, then multiply each fraction’s numerator and denominator by the missing factor of the common denominator.

Remember that multiplication is commutative: that means x • 3 = 3 • x, and x • 5 = 5 • x.

x

3

5

2+ Original expression

x

x x•

3

5

5

5•

2=

+

Multiply 3

5 by

x

x and

x

2 by

5

5.

x

x x

•3

•5

5•2

5•= + Simplify each term.

x

x x

3

5

10

5= + Multiply.

x

x

3 10

5=

+ Rewrite the expression by adding the numerators over the common denominator.

The rewritten expression is x

x

3 10

5

+.


Example 2


5

13

++ .


With an x in the denominator of the first term, x

5

1+, we have to

ensure that x + 1 ≠ 0 in order to keep this expression valid. Set the

denominator equal to 0 and solve this resulting equation for x to

determine values of x that make the expression invalid.

x + 1 = 0

x = –1

An x-value of –1 invalidates the expression, so x ≠ –1.


The second term, 3, does not have an apparent denominator. Yet

any number can be rewritten as a ratio over 1; therefore, the second

term is equivalent to 3

1. Thus, if we rewrite the rational expression as

x

5

1

3

1++ , it can be seen that the two denominators are x + 1 and 1.

Their product, x + 1, is a useful common denominator.

4. Check to see if the result can be written in a simpler form.

The numerator does not have either 5 or x as a factor; therefore, it

cannot be written in any other format that will further simplify the

result. Thus x

x

x

3

5

2 3 10

5+ =

+, where x ≠ 0, is the simplest

way to rewrite the original expression, x

3

5

2+ .



The denominator of the first term is already x + 1, so it doesn’t need to be multiplied by anything.

x

5

1

3

1++ Rewritten expression from the previous step

x

x

x

5

1

3

1•

1

1=

++

++

Multiply 3

1 by

x

x

1

1

++

.

x

x

x

5

1

3( 1)

1=

++

++

Simplify each term.

x

x

5 3( 1)

1=

+ ++

Rewrite the expression by adding the numerators over the common denominator.

x

x

5 3 3

1=

+ ++

Simplify.

x

x

3 8

1=

++


x

3 8

1

++

.


There are no common factors of the numerator or denominator.

This expression cannot be written in any other format that will

simplify the result. Thus x

x

3 8

1

++

, where x ≠ –1, is the simplest way

to rewrite the original expression, x

5

13

++ .


Example 3

Simplify the rational expression x x x

5 3 1

42 + + .


In the first rational term, x2 ≠ 0. Taking the square root of each side results in x ≠ 0. This same value also makes the other two denominators invalid. We only need to avoid this one value, so the domain of the expression is x ≠ 0.


As in previous examples, we can determine a common denominator by finding the product of the denominators. That gives us x2 • x • 4x = 4x4. This would work just fine (it is most certainly a common denominator), however, it does seem large. We can likely determine a smaller common denominator.

Each of the three denominators in the original expression has x as a factor, so we only need to include it once. The first term has an additional factor of x, and the last term has an additional term of 4. The product of these three values—the common x, another x from the first term, and the 4 from the last term—is 4x2. This is a smaller and perhaps easier to handle denominator.

Thus, we can use 4x2 as the common denominator.



x x x

5 3 1

42 + + Original expression

x

x

x x

x

x x

4

4•

5 4

4•

3•

1

42=

+

+

Multiply x

52 by

4

4,

x

3 by

x

x

4

4,

and x

1

4 by

x

x.

x

x

x

x

x

20

4

12

4 42 2 2= + + Simplify each term.

x x

x

20 12

4 2=+ + Rewrite the expression by

adding the numerators over the common denominator.

x

x

20 13

4 2=+

Combine like terms.


x

20 13

4 2

+.


There are no common factors of the numerator or denominator. This

expression cannot be written in any other format that will simplify the

result. Therefore, x

x

20 13

4 2

+, where x ≠ 0, is the simplest way to rewrite

the original expression.

You can use this strategy to combine many different rational expressions, including those that appear to be more complex. The expressions may involve more terms, or perhaps subtraction here and there. Yet the process outlined in this example remains the same. Carefully following these same steps will yield successful results.


Example 4


x x

3

2 1

8

3++

−.


The first rational term has a denominator of 2x + 1. Set the denominator equal to 0 and then solve the resulting equation for x to determine values that make this expression invalid.

2x + 1 = 0

2x = –1

x1

2=−

An x-value of 1

2− invalidates the expression, so x

1

2≠− .

The second rational term has a denominator of x – 3. Again, set the denominator equal to 0 and then solve the resulting equation for x to determine values that make this expression invalid.

x – 3 = 0

x = 3

An x-value of 3 invalidates the expression, so x ≠ 3.

The rational expression x

x x

3

2 1

8

3++

− is only valid when x

1

2≠− and

x ≠ 3.


The two denominators are 2x + 1 and x – 3. They share no factors, so their product will be a good common denominator.

(2x + 1)(x – 3) = 2x(x – 3) + 1(x – 3)

= 2x2 – 6x + x – 3

= 2x2 – 5x – 3

The common denominator is 2x2 – 5x – 3.



x

x x

3

2 1

8

3++

−Original expression

x

x

x

x

x

x x

3

3•

3

2 1

2 1

2 1•

8

3=

−− +

+++ −

Multiply x

x

3

2 1+ by

x

x

3

3

−−

and x

8

3− by

x

x

2 1

2 1

++

.

x x

x x

x

x x

3 ( 3)

( 3)(2 1)

8(2 1)

(2 1)( 3)=

−− +

++

+ −Simplify each term.

x x

x x

x

x x

3 9

2 5 3

16 8

2 5 3

2

2 2=−

− −+

+− −

Multiply.

x x x

x x

3 9 (16 8)

2 5 3

2

2=− + +

− −


x x

x x

3 7 8

2 5 3

2

2=+ +− −

Combine like terms.


We already know the denominator has x – 3 and 2x + 1 as factors. The numerator, however, has no integer factors. This expression cannot be written in any other format that will simplify the result.

Thus x x

x x

3 7 8

2 5 3

2

2

+ +− −

, where x1

2≠− and x ≠ 3, is the simplest way to

rewrite the original expression, x

x x

3

2 1

8

3++

−.


Example 5


x

x x x

1

3

10 6

2 3

4

12−+

−− −

−+

.


There are three terms in the expression and they each have a denominator.

Because it appears to be the most complex, start with the second denominator, x2 – 2x – 3. This expression is quadratic, so try to factor it.

x2 – 2x – 3 = (x – 3)(x + 1)

The factors are (x – 3) and (x + 1). Set each of these equal to 0 and solve for x to determine values that make the expression invalid.

x – 3 = 0 x + 1 = 0

x = 3 x = –1

An x-value of 3 invalidates the expression, as does an x-value of –1. Therefore, we can now more clearly see that x ≠ 3 and x ≠ –1 for the second term’s denominator. That takes care of the third term’s denominator, x + 1, since it is one of the factors of the second denominator; in other words, x ≠ –1 is likewise a restriction of the third term in the rational expression. We can verify this by setting the denominator of the third term equal to 0 and solving for x.

x + 1 = 0

x = –1

Thus, an x-value of –1 invalidates the expression, and x ≠ –1 for the third term’s denominator.

Now we need to determine invalid values for the first term’s denominator, 3 – x.

Because 3 – x = –(x – 3), we can rewrite the first term as follows.

x x

1

3

1

3−=

−−

Now we can see that the first term’s denominator, rewritten as x – 3, is a factor of the second denominator as well. That means that x ≠ 3 and x ≠ –1 for all three terms of the expression.



Because the second term’s denominator, x2 – 2x – 3, has the denominators of the first term and third term as factors, the second term’s denominator is our common denominator.


The middle term needs no adjustment, since it already includes the common denominator.

x

x

x x x

1

3

10 6

2 3

4

12−+

−− −

−+

Original expression

x

x

x

x

x x x

x

x

1

3•

1

1

10 6

2 3

4

1•

3

32=−−

++

+−

− −−

+−−

Multiply

x

1

3

−−

by x

x

1

1

++

and x

4

1+ by

x

x

3

3

−−

.

x

x x

x

x x

x

x x

1( 1)

( 3)( 1)

10 6

2 3

4( 3)

( 1)( 3)2=− +− +

+−

− −−

−+ −

Simplify each term.

x

x x

x

x x

x

x x

1 1

2 3

10 6

2 3

4 12

2 32 2 2=− −− −

+−

− −−

−− −

Multiply.

x x x

x x

( 1 1) (10 6) (4 12)

2 32=− − + − − −

− −


x x x

x x

1 1 10 6 4 12

2 32=− − + − − +

− − Simplify.

x

x x

5 5

2 32=+

− −Combine like terms.



We already know the denominator has x – 3 and x + 1 as factors. The numerator can be factored, because both terms include a factor of 5.

Factor the expression to simplify it even further.

x

x x

5 5

2 32

+− −

Expression from the previous step

x

x x

5( 1)

( 1)( 3)=

++ −

Factor the numerator and denominator.

x

x x

5 ( 1)

( 1) ( 3)=

++ −

Cancel out common factors.

x

5

3=

−Simplify.

The factor x + 1 cancels out of the numerator and the denominator.

Therefore, the simplest way to rewrite the expression

x

x

x x x

1

3

10 6

2 3

4

12−+

−− −

−+

is x

5

3−, where x ≠ 3.

Lesson 1: Operating with Rational ExpressionsU3-15

PRACTICE

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPSLesson 1: Operating with Rational Expressions

For problems 1–7, simplify each rational expression. Whenever possible, reduce the expression to its lowest terms. State any restrictions on x.

1. x

x15

72−

2. x

xx

2 132

++

3. x

x

x

x

5

5

5

5

+−

+−+

4. x

x

8

5

3

1−

+

5. x

xx

3

12 12 +

+ −

6. x

x

x

x x

8

7

7 13

12 352

+−

−−

− +

7. x

x

x x

x

x

5

3 2

2 9

6 19 10

3 2

2 52−−

+− +

++−

Practice 3.1.1: Adding and Subtracting Rational Expressions

continued

Unit 3: Rational and Radical RelationshipsU3-16

PRACTICE


Use your knowledge of rational expressions to complete problems 8–10. State any restrictions on x.

8. From a package containing x

x

1

3

2 −−

cookies, Jodi’s class removes (x + 2) cookies for

quality control experiments. What expression represents the remaining cookies?

9. A triangle has three sides described by the rational expressions x 1− , x

x

3 2+,

and x x

x

5 2 1

2 1

2 + ++

, with all units in centimeters. What simplified expression

describes the perimeter of this triangle?

10. Colby climbs a steep cliff that rises x x

x

2 15

2 5

2 + −−

meters above the beach. On

the way down, he stops to rest at a ledge that is x

x

2

6− meters below the summit.

What simplified rational expression represents his height above the beach when

he is resting on the ledge?


Lesson 3.1.2: Multiplying Rational ExpressionsIntroduction

Like with multiplying fractions, finding the product of rational expressions requires that the values in the numerator be handled separately from the values in the denominator. As the expressions grow more complex, manipulating these values is similar to steps taken previously when working with polynomials. Looking for possible invalid values and opportunities to simplify the expression can make the task less complex.

Key Concepts

• Multiplication of rational expressions does not require a common denominator.

• The product of a collection of rational expressions is the product of the numerators divided by the product of the denominators. Written symbolically,

p xa x

z x

b x

y x

c x

w x

a x b x c x

z x y x w x( )

( )

( )•

( )

( )•

( )

( )

( )• ( )• ( )

( )• ( )• ( )= = .

• As previously encountered, any values that make the denominator equal to 0

must be identified and removed from the domain. For example, the value x =

1 must be excluded from the rational expression x

x

2

1−, since that value makes

the rational expression equivalent to x

x

2

1

2(1)

(1) 1

2

0−=

−= , an undefined form.

• Similar to fractions, rational expressions can (and often should) be reduced by eliminating common factors from both the numerator and denominator.




3

1•8

+.


The denominator of the first term, x + 1, cannot equal 0. Set this expression equal to 0 and then solve for x to determine values that make this expression invalid.

x + 1 = 0

x = –1

An x-value of –1 invalidates the expression, so x ≠ –1.

2. Rewrite the second term as a fraction.

The number 8 can be rewritten in equivalent form as 8

1.

3. Multiply the expressions to form a single rational expression.

x x x

3

1•

8

1

3•8

1( 1)

24

1+=

+=

+

4. Check for any factors that might make it possible to further simplify the resulting expression.

The individual terms from the original expression are both simple. There is no common factor.

The final result of x x

3

1•8

24

1+=

+ is

x x

3

1•8

24

1+=

+ when x ≠ –1.


Example 2


xx

3•5

−.


The denominator of the first term, x, cannot equal 0. Therefore, x ≠ 0.

2. Rewrite the second term as a rational expression.

The expression 5x can be rewritten as x5

1.


x

x

x x x

x

x x

x

3•

5

1

5 ( 3)

•1

5 152−=

−=

−


The numerator shares a factor, x, with the denominator (you might have noticed this in the last step, and adapted then). Factored, the result looks like this:

x x

x

x x

xx

5 15 5 155 15

2 ( )−=

−= −

The final result of x

xx x

3•5 5 15

−= − is

x

xx x

3•5 5 15

−= − when x ≠ 0.


Example 3


xx

5•

12

+.


Both expressions are invalid when x = 0. Therefore, x ≠ 0.


x

x

x

x

x x

x

x

5•

1 5( 1)

•

5 52 2 3

+=

+=

+

3. Check for any common factors that might make it possible to further simplify the resulting expression.

There is no common factor. Though the numerator and denominator both have an x term, x is not a common factor; we cannot factor x out of 5x + 5.


x

x

x

x

5•

1 5 52 3

+=

+ is

x

x

x

x

x

5•

1 5 52 3

+=

+ when x ≠ 0.

Example 4


x

x

x

1•

2 3

12

+ −−

.


The denominator of the first term cannot equal 0. That means x ≠ 0. In the second term, x – 1 ≠ 0, which means x ≠ 1.

Therefore, x ≠ 0 and x ≠ 1.



x

x

x

x

x x

x x

x x

x x

1•

2 3

1

( 1)(2 3)

( 1)

2 32 2

2

3 2

+ −−

=+ −

−=

− −−


There are no common factors.


x

x

x

x x

x x

1•

2 3

1

2 32

2

3 2

+ −−

=− −−

is x

x

x

x

x x

x x

1•

2 3

1

2 32

2

3 2

+ −−

=− −−

when x ≠ 0 and x ≠ 1.

Example 5

Simplify the rational expression x x

x

x

x x

4 5

1•

8

9 8

2

2

− −+

++ +

.


The denominator of the first term means x + 1 ≠ 0, or x ≠ –1.

The second term is a quadratic, and must be factored to determine invalid values for x.

x2 + 9x + 8 = (x + 1)(x + 8)

The factors are (x + 1) and (x + 8). Set each of these equal to 0 and solve for x to determine values that make the expression invalid.

x + 1 = 0 x + 8 = 0

x = –1 x = –8

An x-value of –1 invalidates the expression, as does an x-value of –8. Therefore, for the second denominator, x ≠ –1 and x ≠ –8.

Notice that x ≠ –1 is also a restriction on the denominator of the first term. Thus, for the entire rational expression, x ≠ –1 and x ≠ –8.


2. Factor the numerator of the first term.

Since the numerator of the term x x

x

4 5

1

2 − −+

appears to be quadratic,

factoring this numerator may aid in simplifying the expression.

The numerator of the first term is x2 – 4x – 5.

x2 – 4x – 5 = (x – 5)(x + 1)

Notice that (x + 1) is one of the factors of the denominator of the second term.

3. Using the previously determined factors, multiply the expressions to form a single rational expression.

Replace the quadratic in each term of the original expression with its factored form.

x x

x

x

x x

x x

x

x

x x

4 5

1•

8

9 8

( 5)( 1)

1•

8

( 1)( 8)

2

2

− −+

++ +

=− +

++

+ +

Multiply the rewritten terms.

x x

x

x

x x

x x x

x x x

( 5)( 1)

1•

8

( 1)( 8)

( 5)( 1)( 8)

( 1)( 1)( 8)

− ++

++ +

=− + ++ + +

4. Simplify the rational expression.


x x x

x x x

x x x

x x x

x

x

( 5)( 1)( 8)

( 1)( 1)( 8)

( 5) ( 1) ( 8)

( 1) ( 1) ( 8)

5

1

− + ++ + +

=− + ++ + +

=−+

The final result of x x

x

x

x x

x

x

4 5

1•

8

9 8

5

1

2

2

− −+

++ +

=−+

is x x

x

x

x x

x

x

4 5

1•

8

9 8

5

1

2

2

− −+

++ +

=−+

when

x ≠ –1 and x ≠ –8.


PRACTICE


For problems 1–5, simplify each rational expression. Whenever possible, reduce the expression to its lowest terms. If necessary, factor the terms in each expression before multiplying. State any restrictions on x.

1. (2 5)•2

2x

x

x+

+

2. 3

5•

3 7

4

2x

x

x x

x++ +−

3. 8

2 3•

10 15

7 82

x

x

x

x x

++

++ −

4. 8 9

10•

2

9

2x x

x

x

x

− −−

5. 3

1•

2 3

7 12•

4

1

2

2 2

x

x

x x

x x

x

x

++

− −+ +

++

Practice 3.1.2: Multiplying Rational Expressions

continued


PRACTICE


Use your knowledge of rational expressions and factoring to complete problems 6–10. State any restrictions on x.

6. For what values of x is x x x

x

12

16

3 2

2

− −−

an invalid expression?

7. Show that the product of the expressions x x

x

2

9 3

2 −+

and x

x

3 6

42

+−

is equivalent to

the expression x

x 3+.

8. The net force applied to an object is the product of the object’s mass and

its acceleration. If the expression 12x describes the mass of a plastic ball in

kilograms and the acceleration of the ball is x

x

3 4

4 16

2 ++

meters per second2, what

simplified expression represents the net force in Newtons (N) applied to the ball?

9. A triangle has a height of x x

x

2 1

3

2 + +−

centimeters and a base of x

102 centimeters.

The area of a triangle is equal to 1

2 the product of its base and its height. What

simplified expression represents the area of this triangle?

10. A rectangular box has a length of x

x

4

5

−+

meters, a width of x

x

42

+ meters, and a

height of x x

x

6 5

16

2

2

+ +−

meters. What simplified rational expression represents the

volume of the box?


Lesson 3.1.3: Dividing Rational ExpressionsIntroduction

Previously, explorations of ways to recognize and simplify rational expressions have been used to add, subtract, and multiply those expressions. In this section, these skills will be extended to explore the results of the operation of division on rational expressions.

As in previous examples, familiarity with numerical fractions will be useful in

working with rational expressions. Dividing rational expressions will also be made

easier using a basic fact learned in dealing with fractions; namely, that division is the

equivalent to multiplying by the reciprocal of the divisor. Recall that a reciprocal

is a number that, when multiplied by the original number, has a product of 1. For

example, dividing 12 by 3 is the same as multiplying 12 by the reciprocal of 3, which is 1

3. We can verify that 3 and

1

3 are reciprocals by showing that 3 •

1

3 = 1.

Key Concepts

• The division expression a x b x( ) ( )÷ is equivalent to ( )•1

( )a x

b x: the product

of the divisor and the reciprocal of the dividend.

• A rational expression of the form a x

b x

c x

d x

( )

( )

( )

( )÷ is equivalent to the rational

expression ( )

( )•

1( )

( )

( )

( )•

( )

( )

a x

b x c x

d x

a x

b x

d x

c x= .

• This expression is the product of the divisor and the reciprocal of the dividend.

• A rational expression of the form a x

b x

( )

( ) is equivalent to the expression

a x b x( ) ( )÷ .

• If b(x) is a factor of a(x), there is no remainder.

• If b(x) is not a factor of a(x), then the remainder can be described as r(x).

• If b(x) is not a factor of a(x), a x

b x

( )

( ) is equivalent to the sum of the quotient q(x)

and the ratio r x

b x

( )

( ).


• We can use polynomial division to represent the original rational expression in a different form.

• For example, let a(x) = x2 + 9x + 14 and b(x) = x + 7. If a x

b x

x x

x

( )

( )

9 14

7

2

=+ ++

, then

the rational expression x x

x

9 14

7

2 + ++

is equivalent to the expression (x2 + 9x + 14)

÷ (x + 7). To find the quotient and remainder, use polynomial division:

)x x x

x x

x

x

x7 9 14

7

2 14

2 14

0

22

2

+ + +

+++

+

• In this example, the quotient, q(x), is equal to x + 2, and the remainder, r(x), is

0. The original expression a x

b x

( )

( ) is equivalent to

x x

xx

x

9 14

72

0

7

2

( )+ ++

= + ++

,

or, more simply, x x

xx

9 14

72

2 + ++

= + .

• Synthetic division (a shortcut for polynomial division) can be used to ease the

task of rewriting rational expressions. The rational expression x x

x

8 15

5

2 + ++

results in the following synthetic division.

5 1 8 15

5 151 3 0

−− −

• This result yields a quotient of x + 3, with a remainder of 0. This shows that

the original expression x x

x

8 15

5

2 + ++

is equivalent to xx

( 3)0

5+ +

+, or, more

simply, x + 3.



Simplify the rational expression x x3

7 3

2

÷ .

1. Rewrite the division problem as a multiplication problem, using the reciprocal of the divisor.

The divisor is x

3; the reciprocal of

x

3 is

x

3.

x x3

7 3

2

÷ Original expression

x

x

3

7•

32

=

Multiply by the reciprocal.

The rewritten expression is 3

7•

32x

x.

2. Identify any invalid values of the rewritten expression.

The denominator of the first term is a constant, and is therefore valid. The second term requires that x ≠ 0.

3. Multiply the terms of the rewritten expression.

3

7•

3 3 •3

7•

9

7

2 2 2x

x

x

x

x

x= =

4. Check for any factors that might make it possible to further simplify the resulting expression, and then factor them out.

There is a common factor of x in both the numerator and the denominator.

9

7

9 •

7

9

7

2x

x

x x

x

x= =

The expression x x3

7 3

2

÷ simplifies to x9

7 when x ≠ 0.


Example 2


x x x1

5

12 2+÷

− +.


The divisor is x x

5

12 − +; the reciprocal of

x x

5

12 − + is

x x 1

5

2 − +.

x

x x x1

5

12 2+÷

− +Original expression

x

x

x x

1•

1

52

2

=+

− +



•1

52

2x

x

x x

+− +

.


The denominators are x2 + 1 and 5.

Since adding 1 to the square of a number will always result in a real number value, and since 5 is a constant, there are no real number values that will make either denominator equal 0.

3. Multiply the terms of the rewritten expression.

1•

1

5

( 1)

5( 1) 5 52

2 2

2

3 2

2

x

x

x x x x x

x

x x x

x+− +

=− ++

=− +

+


There are no common factors.

The expression x

x x x1

5

12 2+÷

− + simplifies to

x x x

x5 5

3 2

2

− ++

.


Example 3


x x

x

x

3

6

3

2

2

2

−− −

÷−+

.


The divisor is x

x

3

2

−+

; the reciprocal of x

x

3

2

−+

is x

x

2

3

+−

.

x x

x x

x

x

3

6

3

2

2

2

−− −

÷−+

Original expression

x x

x x

x

x

3

6•

2

3

2

2=−− −

+−



6•

2

3

2

2

x x

x x

x

x

−− −

+−

.


The denominators are x2 – x – 6 and x – 3. Since x2 – x – 6 is a quadratic, see if it can be factored.

x2 – x – 6 = (x – 3)(x + 2)

The factors are x – 3 and x + 2.

Set each factor equal to 0 and solve for x to determine the values that make the expression invalid.

x – 3 = 0 x + 2 = 0x = 3 x = –2

An x-value of 3 invalidates the expression, as does an x-value of –2. Therefore, for the first denominator, x ≠ 3 and x ≠ –2.

Set the denominator of the second term, x – 3, equal to 0 to determine values that make the expression invalid.

x – 3 = 0

x = 3

Therefore, x ≠ 3, which was also a restriction determined for the first expression.

The invalid values of the rewritten expression are x ≠ 3 and x ≠ –2.


3. Check for factors that may cancel, then multiply the terms of the rewritten expression.

In the rewritten expression, 3

6•

2

3

2

2

x x

x x

x

x

−− −

+−

, there are quadratic

expressions in both the numerator and denominator of the first term.

Both of these quadratic expressions can be factored.

We have already determined the factors of x2 – x – 6 are x – 3 and x + 2.

Factor the quadratic in the numerator of the first term, x2 – 3x.

x2 – 3x = x(x – 3)

The factors are x and x – 3.

Notice that in both the numerator and denominator of the first term, x – 3 is a factor.

Use this information to cancel out common factors and multiply the rewritten expression.

3

6•

2

3

2

2

x x

x x

x

x

−− −

+−

Rewritten expression

x x

x x

x

x

( 3)

( 3)( 2)•

2

3=

−− +

+−

Factor the first term of the expression.

x x

x x

x

x

( 3)

( 3) ( 2)•

2

3=

−− +

+−


x

x 3=

−Simplify.

The expression x x

x x

x

x

3

6

3

2

2

2

−− −

÷−+

simplifies to x

x 3−.


There are no additional common factors.

The expression x x

x x

x

x

3

6

3

2

2

2

−− −

÷−+

simplifies to x

x 3−

when x ≠ 3 and x ≠ –2.


Example 4


x x

x x

x x

9 14

7

2

4 4

2

2

2

2

+ ++

÷+ −−

.


The divisor is x x

x x

2

4 4

2

2

+ −−

; the reciprocal of x x

x x

2

4 4

2

2

+ −−

is x x

x x

4 4

2

2

2

−+ −

.

x x

x x

x x

x x

9 14

7

2

4 4

2

2

2

2

+ ++

÷+ −−

Original expression

x x

x x

x x

x x

9 14

7•

4 4

2

2

2

2

2=+ ++

−+ −


The rewritten expression is 9 14

7•

4 4

2

2

2

2

2

x x

x x

x x

x x

+ ++

−+ −

.


The denominators are x2 + 7x and x2 + x – 2. Both of these denominators are quadratic, so see if they can be factored.

x2 + 7x = x(x + 7)

The factors of the denominator of the first term are x and x + 7.


x = 0 x + 7 = 0

x = –7

An x-value of 0 invalidates the expression, as does an x-value of –7. Therefore, for the first denominator, x ≠ 0 and x ≠ –7.

(continued)


Next, factor the second denominator, x2 + x – 2.

x2 + x – 2 = (x + 2)(x – 1)

The factors of the denominator of the second term are x + 2 and x – 1.


x + 2 = 0 x – 1 = 0

x = –2 x = 1

An x-value of –2 invalidates the expression, as does an x-value of 1. Therefore, for the second denominator, x ≠ –2 and x ≠ 1.

The values of x which make the rewritten expression invalid are x ≠ 0, x ≠ –7, x ≠ –2, and x ≠ 1.

3. Check for factors that may cancel, then multiply the terms of the rewritten expression.

In the rewritten expression, 9 14

7•

4 4

2

2

2

2

2

x x

x x

x x

x x

+ ++

−+ −

, there are

quadratic expressions in both the numerator and denominator of

both expressions. Rewrite each quadratic expression in factored form

to determine if there are common factors that may cancel.

We have already determined the factored form of each denominator:

x2 + 7x = x(x + 7) x2 + x – 2 = (x + 2)(x – 1)

Now, factor the numerators.

x2 + 9x + 14 = (x + 2)(x + 7) 4x2 – 4x = 4x(x – 1)

(continued)


Example 5

Use synthetic division to rewrite a x

b x

x x x

x

( )

( )

2 3

1

3 2

=+ − +

+ in the form

a x

b xq x

r x

b x

( )

( )( )

( )

( )= + .

1. Identify any values that make the expression invalid.

If x + 1 = 0, the denominator equals 0; this creates an invalid form. Since x + 1 cannot equal 0, x ≠ –1.

2. Set up the synthetic division problem to divide the given numerator, a(x), by the denominator, b(x).

The numerator is x3 + 2x2 – x + 3. The coefficients of the numerator are 1, 2, –1, and 3. Because we are dividing by x + 1, the number –1 must go on the “shelf ” of the synthetic division. The resulting synthetic division problem is 1 1 2 1 3− − .

Substitute the factored forms into the rewritten expression to cancel out common factors and multiply the terms.

9 14

7•

4 4

2

2

2

2

2

x x

x x

x x

x x

+ ++

−+ −

Rewritten expression from step 1

x x

x x

x x

x x

( 2)( 7)

( 7)•

4 ( 1)

( 2)( 1)=

+ ++

−+ −

Write the quadratic expressions in factored form to identify common factors.

x x

x x

x x

x x

( 2) ( 7)

( 7)•

4 ( 1)

( 2) ( 1)=

+ ++

−+ −


= 4 Simplify.

The expression x x

x x

x x

x x

9 14

7

2

4 4

2

2

2

2

+ ++

÷+ −−

simplifies to 4

when x ≠ 0, x ≠ –7, x ≠ –2, and x ≠ 1.


3. Solve the synthetic division problem.

Perform synthetic division to obtain the following result:

1 1 2 1 3

1 1 2

1 1 2 5

− −− −

−

The first row, 1 1 2 1 3− − , comes from the example as shown in step 2. The next step in solving is to simply bring the first number (1) down to the bottom row. The –1 in the second column is the product of that 1 and –1 (the shelf number). The 1 in the bottom row of the second column is the sum of 2 + (–1). This process is repeated to determine the values in the third and fourth columns. This yields the bottom row: 1 1 2 5− .

4. Identify the remainder.

The remainder is the last entry in the bottom row (shown with a box around it). In this case, the remainder, r(x), is 5.

Since the denominator of the original expression is x + 1, the

remainder of the expression is x

5

1+.

5. Identify the quotient.

The quotient, q(x), is described by the coefficients preceding the remainder in the bottom row. Those coefficients are 1, 1, and –2. Therefore, the quotient is 1x2 + 1x – 2, or x2 + x – 2.

6. Use the results of the synthetic division to rewrite the original expression.

The rational expression x x x

x

2 3

1

3 2+ − ++

simplifies to

x2 + x – 2 + x

5

1+ when x ≠ –1.


PRACTICE


For problems 1–4, simplify each rational expression. Whenever possible, reduce the expression to its lowest terms. State any restrictions on x.

1. x x2 7

5 3

−÷

2. x

x

x3

3

3

9

+−

÷+

3. x x

x

5 1

2

3−÷

+

4. x

x

x

x9

2

3−÷

+−

Use your knowledge of rational expressions to complete problems 5–10. State any restrictions on x.

5. Show that the expression x x

x x

x

x

15 27 6

2

5 1

3 3

2

2

+ −+ −

÷−−

is equivalent to 9.

6. Use synthetic division to rewrite the rational expression x x x

x

3 6 2

2

3 2− + −−

in the

form q xr x

b x( )

( )

( )+ .

Practice 3.1.3: Dividing Rational Expressions

continued


PRACTICE


7. Use polynomial division to rewrite the rational expression x x

x

3 7

1

2 + +−

in the

form q xr x

b x( )

( )

( )+ .

8. One way to determine the time it takes to travel a specified distance is to solve

for the quotient of distance traveled and the average velocity. What expression

represents the time in hours that it will take Mateo to travel a total of

(2x2 – 4x) km at an average velocity, in km per hour, described by the expression x x

x

7 25 12

3

2 − +−

(where x > 3)?

9. Density is the ratio of an object’s mass to its volume. What simplified rational

expression represents the density (in grams per meter3) of an alloy ingot whose

mass in grams is described by the expression x x

x

7 122 + +, with a volume, in

meters3, represented by the expression x x

x

2 8

3

2 + −−

(where x > 3)?

10. On π day (March 14), Tyler’s math class shares x x

x

3 2

1

2 + ++

freshly made pies. If

there are x x

x

8 12

1

2 + +−

students in Tyler’s class who all equally share in the π-day

pies, with x > 1, what expression represents the fraction of pie each student receives?

Lesson 2: Solving Rational and Radical Equations


U3-37Lesson 2: Solving Rational and Radical Equations

WORDS TO KNOW

common denominator a quantity that is a shared multiple of the denominators of two or more fractions

extraneous solution a solution of an equation that arises during the solving process, but which is not a solution of the original equation

least common denominator (LCD)

the least common multiple of the denominators of two or more fractions

proportion a statement of equality between two ratios

radical equation an algebraic equation in which at least one term includes a radical expression

radical expression an expression containing a root

ratio the relation between two quantities; can be expressed in words, fractions, decimals, or as a percentage

rational equation an algebraic equation in which at least one term is expressed as a ratio


MCC9–12.A.CED.1★

MCC9–12.A.REI.2

Essential Questions

1. How can you find a solution to a rational equation?

2. How can you identify extraneous solutions?

3. How can you find a solution to an equation involving radicals?

4. How does the solution to a rational inequality compare to that of a rational equation?


rational inequality a mathematical statement comparing two quantities, including at least one term that is expressed as a ratio

rational number any number that can be written as m

n, where m and n

are integers and n ≠ 0; any number that can be written

as a decimal that ends or repeats


• Math Portal. “Rational Equation Solver.”


This online calculator allows users to solve rational equations. Specific directions and examples for correctly entering data are included.

• Math Warehouse. “How to Solve Radical Equations.”


This site first offers the user a simple algorithm for solving radical expressions. After watching a video that shows the process of solving such an equation, users can practice the solution process with several guided samples.

• Purplemath.com. “Solving Rational Inequalities.”


This site connects solving rational inequalities to solving polynomial inequalities, and provides several worked examples. This resource is particularly helpful for those who have trouble visualizing the graphs or grids necessary to compare the signs of various factors within a larger expression.


Introduction

Your previous experiences have helped you understand how to represent situations

using algebraic equations, and how solving those equations can reveal something

specific about those situations. Most of these equations have involved integer

coefficients. However, coefficients can also be expressed as a ratio. A ratio compares

two quantities, and is often written as a fraction. Numbers expressed as a ratio are

called rational numbers. More formally, a rational number is any number that can

be written as m

n, where m and n are integers and n ≠ 0. In this lesson, we will explore

ways to solve rational equations, and to assess and validate the resulting solutions.

Key Concepts

• A rational equation is an algebraic equation that includes at least one term that is expressed as a ratio.

• A rational equation defines a specific relationship between numbers and variables; this relationship can be analyzed to find a solution to the equation.

• Finding the solution to a rational equation may be accomplished by simplifying the rational terms. To simplify rational terms, convert the rational equation into an equivalent equation with integer coefficients.

• Finding the solution often requires using a common denominator—a quantity that is a shared multiple of the denominators of two or more fractions—to eliminate any rational terms.

• Recall that the least common denominator is the least common multiple of the denominators of the fractions in the expression.

• Like other algebraic equations, a rational equation may have no solutions.

• An extraneous solution can occur when the solution process results in a value that satisfies the original equation, but creates an invalid equation by creating a denominator that is equal to 0.

• An extraneous solution can also occur when a solution results in a value that is irrelevant in the context of the original equation.

Lesson 3.2.1: Creating and Solving Rational Equations


• Recall that a proportion is a statement of equality between two ratios. Techniques for solving proportions can often be used to simplify rational equations.

• The process of solving rational equations can be applied to real-world examples in which a specific task that may be done alone is to be completed cooperatively.



Solve the proportion =x

4

6

8 for x.

1. Determine the least common denominator.

The denominator of the left expression is 4, while the denominator of the right expression is 8.

The least common denominator (LCD) is 8, the smallest multiple with both denominators as factors.

2. Multiply each ratio by the common denominator.

=x

(8)4

(8)6

8Multiply each ratio by the LCD, 8.

=x8

4

48

8Simplify.

3. Simplify the resulting equation.

=x8

4

48

8Equation from the previous step

=x2

1

6

1Simplify the fractions.

2x = 6

The result is an algebraic equation without any remaining rational numbers that can be solved using methods you have already learned.

4. Solve the resulting equation.

2x = 6 Simplified equation from the previous step

x = 3 Divide both sides of the equation by 2.


5. Verify the solution(s).

While it appears that we have a solution, x = 3, we have to return to the original equation to confirm it.

x

4

6

8= Original equation

(3)

4

6

8= Substitute 3 for x.

3

4

6

8=

The result is a true statement; therefore, x = 3 is an acceptable result.

For the proportion =x

4

6

8, x = 3.

Example 2

Solve the rational equation − =t t

5 3

12

9

3 for t.


The denominators in the equation are t, 12, and 3t.

The smallest multiple having all three of these values as a factor is 12t; therefore, the least common denominator is 12t.


2. Multiply each term of the equation by the common denominator and simplify.

− =tt

t tt

(12 )5

(12 )3

12(12 )

9

3Multiply each term by the LCD, 12t.

− =t(12)5

1( )

3

1(4)

9

1Simplify each fraction.

(12)5 – (t)3 = (4)9 Continue to simplify.

60 – 3t = 36

The result is an algebraic equation without any remaining rational numbers that can be solved using methods you have already learned.


60 – 3t = 36 Simplified equation from the previous step

60 = 36 + 3t Add 3t to both sides.

24 = 3t Subtract 36 from both sides.

t = 8 Divide both sides by 3.


While it appears that we have a solution, t = 8, we have to return to the original equation to confirm it. Because there are variables in the denominators of the first and last terms, we must determine what value(s) would make the denominator equal 0, and thereby invalidate the equation (since division by 0 yields an undefined result and is not acceptable).

The expression t

5 is undefined when t = 0; therefore, t ≠ 0.

Similarly, the expression t

9

3 is also undefined when t = 0; once again,

t ≠ 0. Our solution, t = 8, does not violate this condition; therefore, it

is an acceptable result.

For the equation − =t t

5 3

12

9

3, t = 8.


Example 3

Solve the rational equation −+

=y y

2 1

2

1

3 for y.


The denominators are y, y + 2, and 3.

The smallest multiple having all three of these values as a factor is 3y(y + 2), the irreducible product of all three denominators.

The least common denominator is 3y(y + 2).

2. Multiply each term by the common denominator.

+ • − + •+

= + •y yy

y yy

y y3 ( 2)2

3 ( 2)1

23 ( 2)

1

3


Start by rewriting each occurrence of the common denominator in the equation as a ratio:

+ =+

y yy y

3 ( 2)3 ( 2)

1

Common denominator rewritten as a ratio

+• −

+•

+=

+•

y y

y

y y

y

y y3 ( 2)

1

2 3 ( 2)

1

1

2

3 ( 2)

1

1

3Rewritten equation

3(y + 2) • 2 – 3y • 1 = y(y + 2) • 1 Simplify the ratios.

6y + 12 – 3y = y2 + 2yCarry out the multiplication.

3y + 12 = y2 + 2y Subtract 3y from 6y.

The algebraic equation that results this time, 3y + 12 = y2 + 2y, still has only integer coefficients. However, it is no longer linear—this is a quadratic equation because y is raised to a power of 2. We will need to either factor or use the quadratic formula to determine the value of y.


4. Solve the quadratic equation by factoring or using the quadratic formula.

Start by setting the equation equal to 0 so that the equation is in the standard form of a quadratic equation. Then, factor the quadratic.

3y + 12 = y2 + 2y Equation found in the previous step

12 = y2 + (– y) Subtract 3y from both sides.

0 = y2 – y – 12 Subtract 12 from both sides.

0 = (y – 4)(y + 3) Factor the resulting quadratic.

y = 4 or y = –3


Now that we have potential solutions, y = 4 or y = –3, we must return to the original equation. We must determine what value(s) would make the denominators equal 0, and thereby invalidate the equation.

The expression y

2 is undefined when y = 0; therefore, y ≠ 0.

The expression +y

1

2 is also undefined when y = –2; therefore, y ≠ –2.

There is no value for y that invalidates the last term, 1

3, because the

denominator does not contain a variable.

Our solutions, y = 4 or y = –3, do not violate this condition; therefore, they are acceptable results.

If −+

=y y

2 1

2

1

3, then y = 4 or y = –3.


Example 4

Solve the rational equation −

++

=− −p

p

p p p

1

4 2

6

2 82 for p.


The denominators are p – 4, p + 2, and p2 – 2p – 8.

The least common denominator is not apparent with these denominators. However, the product of the first two denominators can be determined.

(p – 4)(p + 2) Product of the first two denominators

p(p + 2) – 4(p + 2)Rewrite the product using the Distributive Property.

p2 + 2p – 4p – 8 Distribute.

p2 – 2p – 8 Simplify.

Since the product of the first two denominators is equal to the third denominator, we have determined a suitable common denominator: p2 – 2p – 8.

2. Multiply each term by the common denominator.

− − •−

+ − − •+

= − − •− −

p pp

p pp

pp p

p p( 2 8)

1

4( 2 8)

2( 2 8)

6

2 82 2 2

2



Think of each term as a ratio, then simplify each ratio by using factors.

− −•

−+

− −•

+=

− −•

− −p p

p

p p p

p

p p

p p

2 8

1

1

4

2 8

1 2

2 8

1

6

2 8

2 2 2

2

− −

•−

+− −

•+

=− −

•− −

p p

p

p p p

p

p p

p p

2 8

1

1

4

2 8

1 2

2 8

1

6

2 8

2 2 2

2

Equation with each term written as a ratio

− +•

−+

− +•

+=

− −•

− −p p

p

p p p

p

p p

p p

( 4)( 2)

1

1

4

( 4)( 2)

1 2

2 8

1

6

2 8

2

2

− +

•−

+− +

•+

=− −

•− −

p p

p

p p p

p

p p

p p

( 4)( 2)

1

1

4

( 4)( 2)

1 2

2 8

1

6

2 8

2

2

Rewritten equation with factors

p p

p

p p p

p

p p

p p

( 4) ( 2)

1

1

4

( 4) ( 2)

1 2

2 8

1

6

2 8

2

2

− +•

−+

− +•

+=

− −•

− −

p p

p

p p p

p

p p

p p

( 4) ( 2)

1

1

4

( 4) ( 2)

1 2

2 8

1

6

2 8

2

2

− +•

−+

− +•

+=

− −•

− −

Divide out like factors.

+• +

−• = •

p pp

( 2)

11

( 4)

11 6 Simplify.

(p + 2)(1) + (p – 4)(p) = 6Simplify the ratios.

p + 2 + p2 – 4p = 6Simplify the equation.

p2 – 3p + 2 = 6Continue to simplify.

p2 – 3p – 4 = 0

The simplified equation, p2 – 3p – 4 = 0, is quadratic.


4. Solve the resulting quadratic equation by factoring or using the quadratic formula.

This quadratic equation can be solved using the quadratic formula,

− ± −b b a c

a

4( )( )

2

2

, where a, b, and c are the coefficients of each term.

− ± −b b a c

a

4( )( )

2

2

Quadratic formula

− − ± − − −( 3) ( 3) 4(1)( 4)

2(1)

2 Substitute 1 for a, –3 for b, and –4 for c.

± +3 9 16

2Simplify.

±3 5

2

p = 4 or p = –1


In the original equation, there are variables in the denominator of each term.

We must determine what value(s) would make those denominators equal 0, and thereby invalidate the equation.

The expression −p

1

4 is undefined when p = 4; therefore, p ≠ 4.

The expression +p

1

2 is also undefined when p = –2; therefore, p ≠ –2.

(continued)


When we found our least common denominator, we determined

that the expression p2 – 2p – 8 = (p – 4)(p + 2). That means

− −=

− +p p p p

1

2 8

1

( 4)( 2)2 .

The same results as before are true: p ≠ 4 and p ≠ –2.

One of our algebraic solutions is that p = 4. This is not a valid solution, because it would result in a 0 in the denominator. This kind of solution is called an extraneous solution, since it satisfies our algebraic problem, but makes no sense in the original equation or problem.

However, our other solution, p = –1, poses no such problem. It is an acceptable result, and therefore is our only solution to the original problem.

If −

++

=− −p

p

p p p

1

4 2

6

2 82 , then p = –1.

Example 5

Demarco and his friend Chase have a small computer business. Demarco can set up a computer for a customer in 90 minutes. Chase can do the same job in 2 hours. Wednesday afternoon, a new client contacts them to set up 14 computers by Friday afternoon. Create a rational equation to find the time it will take for Demarco and Chase to complete this task.

1. Identify the variable.

In this example, the total amount of time it will take Chase and Demarco to complete the task is not known.

Let t describe the total amount of time it will take for the task to be completed.

Notice that there are two time units described in the problem: hours and minutes. Select one unit to use consistently.

Using hours, the work times are 1.5 and 2 hours; using minutes, the work times are 90 and 120 minutes. Because rational expressions and decimals do not always mix well, minutes are a better choice (though the choice will not affect the result).


2. Identify the task that is to be completed.

Demarco and Chase need to set up 14 computers for their client.

3. For each participant in the problem, create a ratio that describes the proportion of the task to be completed.

There are two participants in this problem: Demarco and Chase.

In t minutes, Demarco can set up t

90 computers.

In t minutes, Chase can set up t

120 computers.

4. Create a rational equation to find the time it will take for Demarco and Chase to complete the task together.

Because Demarco and Chase are working together, use the sum of their individual efforts to create an equation.

Demarco’s efforts + Chase’s efforts = completed task

t t

90 12014+ =

The equation t t

90 12014+ = represents the time it will take for

Demarco and Chase to set up 14 computers.

Lesson 2: Solving Rational and Radical EquationsU3-51

PRACTICE

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPSLesson 2: Solving Rational and Radical Equations

For problems 1–3, solve each rational equation for x.

1. x x1

33

5

+− =

2. x x

6 1

4

9

1− =

−

3. x

x

x

x

2

1

5

112−

+−−

=

Use the given information and what you know about rational equations to complete problems 4–10.

4. The sum of the reciprocals of two consecutive, positive even integers is 5

12.

What are the two integers?

5. Tyler can shovel the driveway in 2 hours. Dakota can complete the job in 90 minutes. What rational equation represents the information presented in this problem?

6. Emily spends her weekends cliff climbing. She can ascend a steep 480-foot cliff near her home with little difficulty. She can descend from the cliff at 5 times the rate she can ascend, completing the descent in 3 hours less time than it takes her to climb up. Create a rational equation that describes how long it will take her to climb up and down the cliff.

Practice 3.2.1: Creating and Solving Rational Equations

continued


PRACTICE


7. James and Liu volunteer at the food pantry one Saturday each month, packing boxes of food to deliver to families. Working alone, James can pack 120 boxes in 2 hours. Liu can pack 160 boxes in 4 hours. Create a rational equation to find the time it takes them to work together to pack 600 boxes.

8. Working alone, Erica can thoroughly detail a car in 80 minutes. It takes Alex 2 hours to do the same job. One afternoon, Erica and Alex start detailing a customer’s car together. Half an hour later, Jacki, the new employee, joins them. If the three of them finish 15 minutes later, how long would it have taken Jacki to detail the same car on her own?

9. A developer needs to hire a crew to pave the parking lot of a new shopping plaza. Randolph and Sons Paving can complete the job in 8 days. Mortimer Brothers Paving can do the job in 5 days, but can’t start until 2 days later than Randolph and Sons. The developer is anxious to get the job done, and decides to hire both companies. Randolph and Sons works alone for the first 2 days and then is joined by Mortimer. How many days will it take the two companies to pave the parking lot?

10. Dani spends her summers at her grandparents’ cabin in Canada. The cabin is 42 kilometers upriver from the nearest town. The river has a downstream current of 4 kilometers per hour (km/h). If Dani can canoe from the cabin to town and back in 14 hours, what is her average canoeing speed in still water?


Introduction

Each solution to a rational equation is a single, unique value. However, this is not the case with rational inequalities. A rational inequality is a mathematical statement comparing two quantities, and includes at least one term that is expressed as a ratio. In this section, we will explore the process of solving rational inequalities and determining values that make the inequality true.

Key Concepts

• Solutions to a rational equation, if they exist, are a discrete set of values.

• However, solutions to a rational inequality, if they exist, are a range of values.

• Before solving a rational inequality, the inequality should be converted into a

form comparing a single rational expression to 0, such as a x

b x

( )

( )0< .

• Rational inequalities can be rewritten using previously learned methods of adding and subtracting rational expressions.

• A rational expression will be positive, or greater than 0, when the signs of the numerator and the denominator are the same.

• A rational expression will be negative, or less than 0, when the signs of the numerator and the denominator are not the same.

• Factoring the numerator or denominator can make it easier to identify values that affect the sign of the rational expression.

• Because division by 0 is undefined, check for any values that make the denominator equal to 0. These values must be eliminated from any potential solution.

• Identify the interval of values for which the numerator is positive (or negative). Comparing these to the values for which the denominator is positive (or negative) will identify those values for which the inequality is true.

• The solution to an inequality cannot be checked in the same way that the solution to an equation can be checked, since we have no way of substituting every possible result back into the original inequality. We can get a sense of the accuracy of a solution by selecting a value within the solution interval, and testing it to see if the inequality is true for that value. The result can only give an indication that the inequality is correct; it does not prove the inequality to be true.

Lesson 3.2.2: Creating and Solving Rational Inequalities


Example 1

Solve the rational inequality x 4

30

−< for x.

1. Check for any values of x that would make the rational expression undefined.

If the denominator is equal to 0, then the rational expression will be undefined and so will the inequality.

Because the denominator is a constant, there are no values that are excluded from the domain.

2. Analyze the inequality to identify intervals for the variable x that will make the expression less than 0.

Because the denominator is a constant, the denominator is always positive.

If the result of the numerator is less than 0, then the expression will be less than 0, since a negative number divided by a positive number will result in a negative number (a value less than 0).

Values for x that make the numerator less than 0 (in other words, negative) will satisfy the inequality.

3. Identify intervals that affect the sign of the numerator.

We can determine when the numerator is negative by solving the linear inequality x – 4 < 0 for x.

x – 4 < 0 Numerator set as a linear inequality

x < 4 Add 4 to both sides.

The numerator will be negative if x < 4.

Guided Practice 3.2.2


4. State the possible solution(s).

The possible solution to the inequality x 4

30

−< is x < 4.

5. Check the possible solution(s).

Choose a value that satisfies the possible solution, x < 4.

One value that satisfies the inequality x < 4 is x = 0.

Test 0 in the original inequality to see if it yields a true statement.

x 4

30

−< Original inequality

(0) 4

30

−< Substitute 0 for x.

4

30− < Simplify.

4

3− is less than 0; therefore, the solution x < 4 is likely valid.

The rational inequality x 4

30

−< when x < 4.


Example 2

Solve the rational inequality x

x 10

+> for x.



The denominator is x + 1; therefore, x + 1 ≠ 0.

Subtracting 1 from both sides gives the equivalent statement x ≠ –1 as a limitation on any solution. Therefore, –1 is excluded from the domain.

2. Identify domain intervals that affect the sign of the numerator and denominator.

If the numerator and denominator are both positive, or if they are both negative, then the resulting ratio will also be positive and, therefore, greater than 0.

Determine the intervals for which the original inequality x

x 10

+> is

true. Find the values for which the numerator and denominator each

equal 0, and then use those values to define the intervals where the

signs of the numerator and denominator should be checked.

In this example, the only number that results in a numerator of 0 is 0, while only –1 results in a value of 0 for the denominator.

Therefore, the intervals that affect the sign of the numerator and denominator are x < –1, –1 < x < 0, and x > 0.


3. Use these intervals to create a table of sign changes.

Use the identified intervals (excluding –1, the invalid value for x described in step 1) to create a table that represents the sign changes for the different values of x in the inequality.

IntervalsFactors x < –1 –1 < x < 0 x > 0

x – – +x + 1 – + +

The entries in the table correspond to the sign of the factor in the left column when the value of x falls within the interval listed at the top of the column.

For example, the shaded cell indicates that the factor x + 1 is negative when x < –1.

4. Use the table to identify the intervals that satisfy the conditions described in step 2.

In step 2, we determined that the expression x

x 1+ is positive when

both the numerator and denominator are the same sign.

The table shows that this happens when x < –1 and when x > 0.


The possible solutions to the inequality x

x 10

+> are x < –1 and x > 0.



Choose a value that satisfies the first possible solution, x < –1.

One value that satisfies the inequality x < –1 is x = –2.

Test –2 in the original inequality to see if it yields a true statement.

x

x 10

+> Original inequality

( 2)

( 2) 10

−− +

> Substitute –2 for x.

2

10

−−

> Simplify.

2 > 0

2 is greater than 0; therefore, the solution x < –1 is likely valid.

Choose a value that satisfies the second possible solution, x > 0.

One value that satisfies the inequality x > 0 is x = 2.


x

x 10

+> Original inequality

(2)

(2) 10

+> Substitute 2 for x.

2

30> Simplify.

2

3 is greater than 0; therefore, the solution x > 0 is likely valid.

The rational inequality x

x 10

+> when x < –1 or x > 0.


Example 3


x

3

402

+−

≥ for x.



The denominator is a quadratic equation, and can be factored as x2 – 4 = (x + 2)(x – 2).

Therefore, the zeros of the quadratic equation are 2 and –2.

These are the values that make the rational expression undefined; in other words, any solution cannot contain ±2.

2. Since the inequality includes 0 as a possible solution, identify any values that make the rational expression equal to 0.

If the numerator is equal to 0, the expression will be equal to 0.

This occurs when x + 3 = 0, or when x = –3.

3. Identify domain intervals that affect the sign of the numerator and denominator.

If the numerator and denominator are both positive, or if they are both negative, then the resulting ratio will also be positive and, therefore, greater than 0.

Determine the intervals for which the original inequality x

x

3

402

+−

≥ is

true. Find the zeros of the numerator and denominator, and then use

those values to define the possible intervals.

In this example, we previously identified –3 as the only zero for the numerator, and ±2 as the zeros for the denominator.

Therefore, the intervals are x ≤ –3, –3 ≤ x < –2, –2 < x < 2, and x > 2.



The intervals found in steps 3 and 4 (excluding ±2, the invalid values for x described in step 1) result in the following table.

IntervalsFactors x ≤ –3 –3 ≤ x < –2 –2 < x < 2 x > 2

x + 3 (numerator)

– + + +

x + 2 – – + +x – 2 – – – +x2 – 4

(denominator)+ + – +

As before, the entries in the table correspond to the sign of the factor in the left column when the value of x falls within the interval listed at the top of the column.

The bottom row represents not a single factor, but the denominator, which is the product of the factors (x – 2) and (x + 2).


We have already determined that the expression x

x

3

42

+−

is positive

when both numerator and denominator are the same sign.

Inspect the table from the previous step. The top row of the table shows the signs of the numerator for various intervals, while the bottom row shows the signs of the denominator.

The table shows that both the numerator and denominator have the same sign when –3 ≤ x < –2 or when x > 2.


The possible solutions to the inequality x

x

3

402

+−

≥ are –3 ≤ x < –2 and x > 2.



Choose a value that satisfies the first possible solution, –3 ≤ x < –2.

One value that satisfies the inequality –3 ≤ x < –2 is x = –3.


x

x

3

402

+−

≥ Original inequality

( 3) 3

( 3) 402

− +− −

≥ Substitute –3 for x.

0

50≥ Simplify.

0 ≥ 0

0 is greater than or equal to 0; therefore, the possible solution –3 ≤ x < –2 is likely valid.

Choose a value that satisfies the second possible solution, x > 2.

One value that satisfies the inequality x > 2 is x = 4.


x

x

3

402

+−

≥ Original inequality

(4) 3

(4) 402

+−

≥ Substitute 4 for x.

7

120≥ Simplify.

7

12 is greater than 0; therefore, the possible solution is likely valid.


x

3

402

+−

≥ when –3 ≤ x < –2 or x > 2.


Example 4


x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

for x.


If the denominator is equal to 0, then the rational expression will be undefined and so will the inequality. Since there are two rational expressions, both of the denominators need to be examined for values that would result in a denominator of 0.

In the first rational expression, x

x 7+, x = –7 would result in a value of

0 for the denominator. Therefore, this value must be excluded from

the solution.

The second rational expression’s denominator, x2 + 8x + 7, is a quadratic equation. It must be factored.

Factoring x2 + 8x + 7 results in (x + 7)(x + 1).

The zeros of this equation are –7 and –1. Therefore, the values x = –7 and x = –1 must be excluded from the solution.

2. Combine the rational expressions on one side of the inequality, leaving 0 on the other.

To get 0 on the right side of the inequality, subtract x x

x x

2 2

8 7

2

2

− −+ +

from

both sides of the inequality.

x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

Original inequality

x

x

x x

x x7

2 2

8 70

2

2+−

− −+ +

≤ Subtract x x

x x

2 2

8 7

2

2

− −+ +

from both sides.

x

x

x x

x x+−

− −+ +

≤7

2 2

( 7)( 1)0

2Rewrite the quadratic denominator in factored form.

x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

can be rewritten as x

x

x x

x x+−

− −+ +

≤7

2 2

( 7)( 1)0

2

.

(continued)


In order to carry out the subtraction of the rewritten expression, a common denominator is needed.

Since the denominator of the first term is also a factor in the denominator of the second term, the common denominator is (x + 7)(x + 1).

x

x

x x

x x+−

− −+ +

≤7

2 2

( 7)( 1)0

2

Rewritten expression

x

x

x

x

x x

x x7•

1

1

2 2

( 7)( 1)0

2

+++

−

− −+ +

≤Multiply the expression

x

x 7+

by x

x

1

1

++

.

x x

x x

x x

x x

( 1)

( 7)( 1)

2 2

( 7)( 1)0

2++ +

−− −+ +

≤ Simplify.

x x

x x

x x

x x( 7)( 1)

2 2

( 7)( 1)0

2 2++ +

−− −+ +

≤ Distribute.

x x x x

x x

( ) (2 2 )

( 7)( 1)0

2 2+ − − −+ +

≤ Rewrite the numerators over the common denominator.

x x

x x

2 3 2

( 7)( 1)0

2 + −+ +

≤ Simplify.

The original inequality x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

can be rewritten as x x

x x

2 3 2

( 7)( 1)0

2 + −+ +

≤ .

3. Since the inequality includes 0 as a possible solution, identify any values that make the rational expression equal to 0.

If the numerator equals 0, the entire expression will be equal to 0.

The numerator, 2x2 + 3x – 2, is a quadratic and can be written in factored form as (x + 2)(2x – 1).

This quadratic has zeros when x = –2 or when x1

2= .

At these values, the rational expression equals 0.


4. Identify intervals that affect the sign of the numerator and denominator.

If the numerator and denominator have different signs (that is, one is positive and the other is negative), then the resulting ratio will be negative and, therefore, less than 0.

Determine the intervals for which this is true. Find the zeros of the

numerator and denominator, and then use those values to define

the possible interval(s) of numbers. We previously identified x = –2

and x1

2= as zeros for the numerator. In step 1, we found that the

denominator has the zeros –7 and –1.

Therefore the intervals are x < –7, –7 < x ≤ –2, –2 ≤ x < –1, x11

2− < ≤ ,

and x1

2≤ .


The intervals found in the previous step (excluding –7 and –1, the invalid values for x described in step 1) result in the following table.

Intervals

Factors x < –7 –7 < x ≤ –2 –2 ≤ x < –1 x< ≤–11

2x≤

1

2x + 2 – – + + +

2x – 1 – – – – +2x2 + 3x – 2

(numerator) + + – – +

x + 7 – + + + +x + 1 – – – + +

(x + 7)(x + 1) (denominator) + – – + +

As before, the entries in the table correspond to the sign of the factor in the left column when the value of x falls within the interval listed at the top of the column. The row marked “numerator” indicates the sign of the product of the numerator’s two factors, and the row marked “denominator” indicates the sign of the product of the denominator’s two factors.



In step 4, we determined that the expression x x

x x

2 3 2

( 7)( 1)

2 + −+ +

is negative

when the numerator and denominator have different signs.

The table shows that the numerator and denominator have opposite

signs when –7 < x ≤ –2 and when x11

2− < ≤ .


The possible solutions to x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

are –7 < x ≤ –2 and

x–11

2< ≤ .


Choose a value that satisfies the first possible solution, –7 < x ≤ –2.

One value that satisfies the inequality –7 < x ≤ –2 is x = –5.


x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

Original inequality

( 5)

( 5) 7

2 2( 5) ( 5)

( 5) 8( 5) 7

2

2

−− +

≤− − − −− + − +

Substitute –5 for x.

5

2

13

8

−≤−−

Simplify.

20

8

13

8− ≤ Rewrite the terms with a

common denominator.

20

8− is less than or equal to

13

8; therefore, the possible solution

–7 < x ≤ –2 is likely valid.(continued)


Choose a value that satisfies the second possible solution, x–11

2< ≤ .

One value that satisfies the inequality x–11

2< ≤ is x = 0.


x

x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

Original inequality

(0)

(0) 7

2 2(0) (0)

(0) 8(0) 7

2

2+≤

− −+ +

Substitute 0 for x.

0

7

2

7≤ Simplify.

02

7≤

0 is less than or equal to 2

7; therefore, the possible solution

x–11

2< ≤ is likely valid.


x

x x

x x7

2 2

8 7

2

2+≤

− −+ +

when

–7 < x ≤ –2 or x–11

2< ≤ .


Example 5

On Earth, the height h above ground of a projectile launched upward can be found using the equation h = h0 + vt – 16t2, where t is the time in seconds after the object is launched, h0 is the height above ground at the time of launch, and v represents the upward velocity (in feet per second) with which the object is launched. The average velocity is the ratio of the distance (in this case, height) and time.

Ariana launched a model rocket from the deck behind her house. The deck is 12 ft high, and the rocket had an initial (launch) velocity of 50 ft/s. Create a rational inequality that can be solved to determine when the average velocity of the rocket is less than 16 ft/s.

1. Identify a general rational expression that represents the given data.

This problem includes a polynomial expression for the height of an object launched upward, and describes the average velocity as a ratio of that velocity and time.

2. Use data from the problem to create a specific rational expression.

The problem has already defined the variable t to represent time, and the equation to represent the height above ground as h = h0 + vt – 16t2.

Since the average velocity is the ratio of those values, then the average

velocity = h

t.

Use this information and the values given in the problem statement to write a rational expression for the average velocity of the rocket.

h vt t

taveragevelocity

1602( )

=+ −

Substitute h0 + vt – 16t2 for h.

t t

taveragevelocity

12 50 16 2( ) ( )=

+ − Substitute 12 for h0 and 50 for v.

The expression that represents the average velocity of the rocket is t t

t

12 50 16 2+ −.


3. Identify the value to which this expression will be compared.

The problem asks for values for which the rocket’s average velocity is less than 16 ft/s.

4. Write the resulting rational expression.

Comparing the rational expression t t

t

12 50 16 2+ − to an average

velocity that is less than 16 ft/s results in the following expression.

t t

t

12 50 1616

2+ −<

t t

t

12 50 1616

2+ −< represents a rational inequality that can be solved

to determine when the average velocity of the rocket is less than

16 ft/s.


PRACTICE


For problems 1–4, solve each of the given rational inequalities.

1. x

8

50

−<

2. x

x

2

13

+≥

3. x

xx

2 17

15

++

< +

4. x

x x x

2 4

2

2

22

−− −

≤−

For problems 5 and 6, use the given information to create a rational inequality for each scenario.

5. Due to the moon’s lower gravitational pull, a projectile launched on the moon would follow a different path than if the projectile were launched from Earth. On the moon, the height h above ground of a projectile launched upward can be found using the equation h = h0 + vt – 2.7t2, where t is the time (in seconds) after the projectile is launched, h0 is the height above ground at the time of launch, and v represents the upward velocity (in meters per second) with which the object is launched. A projectile is launched from a height of 4 ft with an initial (launch) velocity of 25 ft/s. Create a rational inequality that can be solved to determine when the average velocity is less than 30 ft/s.

6. The volume of a pyramid is found using the equation V Bh1

3= , where B is the

area of the base and h is the height. A pyramid is to be designed with a square

base and a volume of 4,000 m3, with a height that is less than the length of one

side of the base. Create an inequality using x that could be solved to find the

possible values of the length of the base.

Practice 3.2.2: Creating and Solving Rational Inequalities

continued


PRACTICE


Use the given information and what you know about rational inequalities to solve problems 7–10.

7. Bobbi and her brother Bentley rake leaves every fall to earn a little extra money. Bobbi can rake all the leaves at one house in 2 hours. Bentley can rake the leaves at one house in 3 hours. There are only 6 houses on the cul-de-sac where they live, but each house has a yard that needs to be raked. If Bobbi and Bentley work together, what is the minimum time they will need to rake all the yards in their cul-de-sac?

8. The expression 9x + 54 represents the cost in dollars for Lenny’s Lumber to produce a sawhorse. How many sawhorses must be built to keep the average cost below $12?

9. The expression t

t

0.06 0.04

4 2

−+

represents the concentration (in g/m2) of

fertilizer sprayed on a suburban lawn, where t is measured in days. When the

concentration exceeds 0.01 g/m2, the fertilizer can be harmful to pets that walk

unprotected over the surface, so signs have to be posted. How many days must

pass after spraying a lawn with fertilizer before such safety signs can be removed?

10. When a new car model is released, sales tend to be climb rapidly with the

excitement (and advertising), before slowing down. Angel, the sales manager at a

local dealership, uses a monthly sales model, t

t

18 21

122

++

, to predict new car model

sales. The variable t represents the number of months since the car’s release. For

what values of t can Angel expect to sell at least 3 of a particular model car?


Introduction

Using the Pythagorean Theorem to solve problems provides a familiar example of a relationship between variables that involve radicals (or roots). Recall that the word root in mathematics has several definitions. You have seen and worked with function roots, or zeros; in this lesson, you will work with a different kind of root—that is, square roots, cube roots, and other radicals. Radical expressions, or expressions containing a root, follow rules that may seem more complicated than those of more common numbers. While you can easily add the numbers 4 and 9 to get 13, there is no similar simple way to add the radicals 4 and 9 without simplifying each individual term first. Since radical expressions require more complex methods in order to be simplified, radical equations also involve more complex methods in order to be solved. However, some common real-life situations can be more readily solved by applying radical equations, making fluency with radical equations a useful skill to learn. In this lesson, we will explore ways to solve radical equations, as well as how to assess and validate the resulting solutions.

Key Concepts

• A radical equation is an algebraic equation in which at least one term includes a radical expression.

• A radical equation defines a specific relationship between numbers and variables that can be analyzed to find a solution.

• It is possible to find the solution to a radical equation by isolating the radical, then raising the expression to an appropriate power to eliminate the radical.

• Like other algebraic equations, a radical equation may have no solutions.

• Recall that an extraneous solution can occur when the solution process results in a value that satisfies the original equation, but creates an invalid equation; in the case of radical equations, extraneous solutions occur when a negative value results under the radical. An extraneous solution can also occur when a solution results in a value that is irrelevant in the context of the original equation.

Lesson 3.2.3: Solving Radical Equations


Example 1

Solve the radical equation − =x 5 2 for x.

1. Isolate the radical expression.

The radical expression ( )−x 5 is already isolated on one side of the equation.

2. Raise both sides of the equation to a power to eliminate the radical.

Because we have a square root on the left side of the equation, we need to raise both sides to the power of 2 to eliminate the radical expression. In other words, square both sides.

− =x 5 2 Original equation

( )− =x 5 22 2 Square both sides of the equation.

x – 5 = 4 Simplify.


x – 5 = 4 Equation found in the previous step

x = 9 Add 5 to both sides.

The solution appears to be x = 9. However, we need to confirm that this is not an extraneous solution.


Substitute the potential solution into the original equation, and check the result.


− =(9) 5 2 Substitute 9 for x.

=4 2 Simplify.

The solution satisfies the original equation, and is therefore valid.

The solution to the radical equation − =x 5 2 is x = 9.



Example 2

Solve the radical equation − =x 5 2 for x.


At first glance, this looks much like the equation given in Example 1. However, “– 5” is not under the radical. This small change creates a big difference, for in this example, the radical expression ( )x is not already isolated on one side of the equation. The first step is to isolate the radical.


=x 7 Add 5 to both sides.

2. Raise both sides of the equation to a power to eliminate the radical, then simplify the result.

Square both sides of the equation to eliminate the radical expression, then solve the equation algebraically.

=x 7 Equation found in the previous step

( ) =x 72 2 Square both sides.

x = 49 Simplify.

The solution appears to be x = 49.




− =(49) 5 2 Substitute 49 for x.

7 – 5 = 2 Simplify.

2 = 2

Since the solution satisfies the original equation, the solution is valid.The solution to the radical equation − =x 5 2 is x = 49.


Example 3

Solve the radical equation − =x x2 for x.


The radical expression ( )− x2 is already isolated on one side of the equation.


Square both sides of the equation to eliminate the radical expression. Solve one side of the equation for 0.

− =x x2 Original equation

( )− =x x2 ( )2 2 Square both sides.

2 – x = x2 Simplify.

2 = x2 + x Add x to both sides.

0 = x2 + x – 2 Subtract 2 from both sides.

The result is a quadratic equation.

3. Solve for x by factoring, or by using the quadratic formula.

The quadratic appears to be factorable.

0 = x2 + x – 2

0 = (x – 1)(x + 2)

The solutions appear to be x = –2 or x = 1.



Substitute the potential solutions into the original equation, and check the result.

Let’s start with x = –2.


− − = −2 ( 2) ( 2) Substitute –2 for x.

=−4 2 Simplify.

2 = –2

The result, 2 = –2, is a false statement. This solution does not satisfy the original equation. –2 is an example of an extraneous solution, one that results from the process of our solution, but does not satisfy the original equation.

Check x = 1.


− =2 (1) (1) Substitute 1 for x.

=1 1 Simplify.

1 = 1

The result is a true statement; therefore, this solution satisfies the original equation.

Since only the second solution satisfies the original equation, only that solution is valid.

The solution to the radical equation − =x x2 is x = 1.


Example 4

Solve the radical equation − =x 7 3 2 for x.

1. Isolate the radical expressions.

The equation contains two distinct radical expressions. Each radical expression is alone on one side of the equation, so it should be possible to eliminate the radicals using the methods previously demonstrated.

2. Raise both sides of the equation to a power to eliminate the radicals, then simplify the result.

Square both sides of the equation, then solve algebraically.

− =x 7 3 2 Original equation

( ) ( )− =x 7 3 22 2

Square both sides.

x – 7 = 9(2) Simplify.

x – 7 = 18

x = 25

The solution appears to be x = 25.



− =x 7 3 2 Original equation

− =(25) 7 3 2 Substitute 25 for x.

=18 3 2 Simplify.

• =9 2 3 2

=3 2 3 2

Since the solution satisfies the original equation, the solution is valid.The solution to the radical equation − =x 7 3 2 is x = 25.


Example 5

Solve the radical equation + = +x x5 3 5 for x.

1. Isolate the radical expressions.

With three different radical expressions, this equation offers no simple path to isolating the radicals. It is not possible to get each radical on one side of the equation because there just aren’t enough sides.

In its current format, the most complex radical is alone on one side. The other two simpler radicals are on the other side. While this radical equation is not as ideally isolated as those in the previous examples, it is as isolated as we need it to be.

2. Raise both sides of the equation to a power to eliminate the radicals, then simplify the result.

Square both sides of the equation, then simplify as much as possible.

+ = +x x5 3 5 Original equation

( ) ( )+ = +x x5 3 52 2

Square both sides.

( )( )+ + = +x x x5 5 3 5 Simplify.

( ) ( )+ + + = +x x x x5 5 5 3 5 Distribute.

( ) ( ) ( ) ( )+ + + = +x x x x x5 5 5 5 3 5Write each term as the product of the distribution.

+ + + = +x x x x5 5 5 3 5 Simplify.

=x x2 5 2

=x x5

While we may not have solved the equation yet, we have managed to rewrite it in a simpler format that we can solve algebraically using previously learned processes.


3. Eliminate the radical from the simplified equation.

As in the previous step, square both sides of the equation, then simplify the result.

=x x5 Simplified equation from the previous step

( ) =x x52 2 Square both sides.

5x = x2 Simplify.

0 = x2 – 5x Subtract 5x from both sides.

0 = x(x – 5) Factor the resulting quadratic.

We can determine the potential solutions by setting each factor of the equation 0 = x(x – 5) equal to 0:

x = 0

x – 5 = 0

Our two potential solutions are x = 0 and x = 5.



Substitute the potential solutions into the original equation, and check the result.

Let’s start with x = 0.


+ = +5 (0) 3(0) 5 Substitute 0 for x.

+ = +5 0 0 5 Simplify.

=5 5

The result, =5 5 , is a true statement. The solution x = 0 appears to be valid.

Now check x = 5.


+ = +5 (5) 3(5) 5 Substitute 5 for x.

+ = +5 5 15 5 Simplify.

=2 5 20

= •

=

2 5 4 5

2 5 2 5

The result, =2 5 2 5 , is a true statement. The solution x = 5 also appears to be valid.

Both potential solutions satisfy the original equation; therefore, both are valid.

The solutions to the radical equation + = +x x5 3 5 are x = 0 and x = 5.


Example 6

Solve the radical equation + + =x3 6 10 73 for x.


Not every radical represents a square root. Higher roots are represented with a radical symbol and a root on the “shelf ” of the radical symbol. The solution process, however, is essentially the same.

+ + =x3 6 10 73 Original equation

+ =−x3 6 33 Subtract 10 from both sides.


Eliminating the radical this time requires raising each side to the third power (i.e., cubing each side), rather than squaring each side.

( )+ = −x3 6 ( 3)33 3 Cube both sides.

3x + 6 = –27 Simplify.3x = –33x = –11

The solution appears to be x = –11.



+ + =x3 6 10 73 Original equation

− + + =3( 11) 6 10 73 Substitute –11 for x.

− + =27 10 73 Simplify.

–3 + 10 = 77 = 7

Since the solution satisfies the original equation, the solution is valid.

The solution to the radical equation + + =x3 6 10 73 is x = –11.


PRACTICE


For problems 1–3, solve each radical equation for x. Round answers to the nearest thousandth.

1. + =−x 7 1

2. + = −x x3 12 2

3. + + =x2 2 2 9 53

Use the given information and what you know about radical equations to solve problems 4–10. Round answers to the nearest thousandth.

4. A right triangle has a hypotenuse measuring 205 mm. If one leg measures 133 mm, what is the length of the other leg?

5. If you attach a ball to a piece of string and whirl it over your head, a tension force

in the string will resist the ball’s tendency to fly away from you. The magnitude of

that force depends on the mass of the ball, the length of the string, and the speed

at which you move the ball. This relationship is described by the formula for

force: =Fmv

r

2

. The mass m of the ball is measured in kilograms, and the length

of the string is r (meters), since the string serves as the radius of a circle. If the

velocity, v, is then measured in meters per second, the force, F, will be in Newtons,

a commonly used metric unit of force represented by the letter N. With what

velocity would you be spinning a 1.25-kilogram ball if the force on a 1.5-meter-

long string is 4.8 N?

Practice 3.2.3: Solving Radical Equations

continued


PRACTICE


6. As a sailboat hull pushes the water aside, a wave forms along the hull. If that wave becomes too big, it pulls back on the boat, slowing it down. There is a maximum speed that any hull can maintain before the force exerted by the wave becomes too large to overcome, and the boat capsizes. This maximum speed can be determined using the formula =v L1.34 , where v represents the sailboat’s velocity through the water in knots (nautical miles per hour) and L is the length along the waterline where the boat and the water surface are in contact. If a team of designers wants their newest sailboat to be able to sail at least 6.25 knots, what is the minimum waterline length the boat can have?

7. The velocity, v, at which a falling object hits the ground (without air resistance) can be found using the formula =v gh2 , where v is the velocity of the object in meters per second (m/s), h is its height in meters, and g is the object’s acceleration due to gravity (in m/s2). Find an object’s acceleration due to gravity if it falls 40 meters and hits the ground at a velocity of 28 m/s.

8. The volume, V, of a pyramid with a square base is related to the height, h, of

the pyramid, and the length of one side of the square base, s, by the formula

=sV

h

3. Find the length of one side of the square base of a pyramid that is

5 cm high and has a total volume of 540 cm2.

continued


PRACTICE


9. The amount of time it takes a pendulum to swing back and forth along its path

is known as its period, and is often symbolized by the lowercase Greek letter tau,

τ. The value of τ is dependent on the pendulum’s acceleration due to gravity, g,

and the pendulum’s length, L, and is found using the formula τ π=L

g2 .

A large science and industry museum features a multi-story pendulum that has

a period of 9 seconds. That is, the pendulum takes 9 seconds to swing from one

end of its arc to the other. If the pendulum’s acceleration due to gravity, g, is

9.8 m/s2, how many meters long is the museum’s pendulum?

10. The amount of time it takes a planet to complete one orbit around its central

star is called a period of revolution, τ. The period of revolution depends on the

planet’s distance from its star, measured in astronomical units (AU); 1 AU ≈

93 million miles (the mean distance between the Earth and the Sun). Kepler’s

Third Law of Planetary Motion quantifies this relationship, stating that the ratio

of the square of the period of revolution and the cube of the planet’s distance

from its star, r, is a constant. Venus has an orbital period of just 0.615 years,

and orbits the Sun at a distance of 0.72 AU. If Mars has an orbital period of

1.88 years, how far is Mars from the Sun? Use the proportion τ τ( )( )

( )( )

=r r

V

V

M

M

2

3

2

3 ,

where the subscripts V and M represent the planets Venus and Mars, to find the

answer.

Lesson 3: Graphing Rational Functions

Unit 3: Rational and Radical Relationships


U3-84



MCC9–12.F.IF.4★

MCC9–12.F.IF.5

MCC9–12.F.IF.7d (+)

MCC9–12.A.REI.11★

Essential Questions

1. How can you determine from a data set whether an equation relating two variables can be written as a rational equation?

2. What are the characteristics of the graph of a rational function and what is the significance of each characteristic?

3. How do you calculate the zeros for a rational function?

4. How can a system of two rational equations have no common solution?

WORDS TO KNOW

asymptote a line that a function gets closer and closer to, but never crosses or touches

degree the greatest exponent attached to the variables in the polynomial

domain the set of all input values (x-values) that satisfy the given function without restriction

extreme value a point on the graph of a rational function that occurs at or near the endpoints of its domain or range

factor one of two or more numbers or expressions that when multiplied produce a given product

factoring the process of separating an expression into two or more prime factors

U3-85Lesson 3: Graphing Rational Functions

horizontal asymptote a straight line of the form y = a which occurs at the maximum or minimum of the graph of a rational function; the graph of the rational function approaches this line but never touches it

quadratic formula a formula that states that the solutions of a quadratic equation of the form ax2 + bx + c = 0 are given by

xb b ac

a=− ± − 4

2

2

range the set of all outputs of a function; the set of y-values that are valid for the function

rational function a function that can be written in the form f xp x

q x=( )

( )

( ),

where p(x) and q(x) are polynomials and q(x) ≠ 0

real zero a solution to an equation that can be graphed on a coordinate plane of real numbers

root the x-intercept of a function; also known as zero

slant asymptote an asymptote of a graph of the form y = ax + b, where a ≠ 0

solution(s) to a system for a system of two equations f(x) and g(x), the value(s) of x for which f(x) = g(x) and that are in the domains of the functions; on a graph, the solutions are the points at which the equations intersect

vertical asymptote a straight line of the form x = b which the graph of a rational function approaches but never touches

zero the x-intercept of a function; also known as root



• Coolmath.com. “Graphing Rational Functions.”


This site includes accessible but extensive coverage of asymptotes, including slant asymptotes.

• Mathway. “Math Problem Solver.”


This graphing utility allows users to plot a variety of mathematical functions, including rational functions. Users can choose from a variety of mathematical symbolic-manipulation tools and a pop-up menu of activities to perform with functions, such as finding zeros.

• Purplemath.com. “Graphing Rational Functions: Introduction.”


This site provides a detailed explanation of how to graph rational functions, with graphing calculator simulations that are helpful for visual learners.

• ThatTutorGuy. “Asymptotes and Rational Functions.”


This site offers streaming video for instruction on horizontal, vertical, and slant asymptotes, as well as examples of how to graph rational functions.


Lesson 3.3.1: Creating Rational Equations in Two VariablesIntroduction

Though we may be unaware of the role they play, rational functions are an important part of daily life. Picture a patient in a doctor’s office who needs to be put on blood pressure medication. How does the doctor know the correct dosage of medication, and how frequently it should be taken, for that person? Or, imagine a factory that manufactures bicycles. How does the production manager determine the optimal amount of bicycles to produce that will yield the highest amount of profit and yet minimize production costs? Across many fields, rational functions are used regularly to help make important decisions.

Key Concepts

• A rational function is the quotient of two polynomial functions.

• In mathematical terms, the rational function f(x) is defined as f xp x

q x=( )

( )

( ),

where p(x) and q(x) are polynomial functions and q(x) ≠ 0.

• For example, the rational function f xx

x=

−−

( )2 1

42 consists of the polynomial functions g(x) = 2x – 1 and h(x) = x2 – 4.

• Recall that the domain of a function is the set of all input values (x-values) that satisfy the given function without restriction.

• The degree of a polynomial is the greatest exponent attached to the variables in the polynomial. The domain of a first-degree rational function is found by identifying the asymptotes.

• An asymptote is a line that a function gets closer and closer to, but never crosses or touches.

• A vertical asymptote is a straight line of the form x = b which the graph of a rational function approaches but never touches.

• A horizontal asymptote is a straight line of the form y = a that occurs at the maximum or minimum of the graph of a rational function.

• The domain will contain every real number except for the point at which the vertical asymptote falls.

• Recall that the domain of a function can be expressed in interval notation. That is, the domain is written in the form of (a, b), where a and b are the endpoints of the interval. Depending on the values of the interval, the notation may change.


• The table that follows summarizes how to note the intervals for various situations. Notice that a parenthesis next to a number indicates that the number is not included in the solution set; however, a bracket indicates that the number is part of the solution set.

Interval notation Example Description

(a, b) (2, 10)All numbers between 2 and 10; endpoints are not included.

[a, b] [2, 10]All numbers between 2 and 10; endpoints are included.

(a, b] (2, 10]All numbers between 2 and 10; 2 is not included, but 10 is included.

[a, b) [2, 10)All numbers between 2 and 10; 2 is included, but 10 is not included.

• The zeros of a rational function are the x-intercepts of the function; in other words, the domain value(s) for which f(x) = 0. Zeros are also known as roots.

• In the example f xx

x=

−−

( )2 1

42 , the domain values are found by assuming

that the denominator x2 – 4 ≠ 0. This means that the numerator 2x – 1 = 0 if

f(x) = 0. Therefore, the zero for this function is found by solving the equation

2x – 1 = 0 for x. The result is x =1

2. Thus, the graph intersects the x-axis at the

point

1

2,0 .

• The values of the domain for which a rational function is undefined are found

by determining the values of the domain for which the denominator is equal

to 0. In the example f xx

x=

−−

( )2 1

42 , the values of the domain for which the

function is undefined are found by solving the equation x2 – 4 = 0 for x, which

gives x = ±2. Therefore, the function is undefined at the domain values of x = 2

and x = –2.


• The values of the domain at which a rational function is undefined are also the

equations of its vertical asymptotes. The vertical asymptotes do not include the

points that make the function true. In the example, f xx

x=

−−

( )2 1

42 , the asymptotes

are x = 2 and x = –2, because =−−

=(2)2(2) 1

(2) 4

3

02f , which is undefined, and

f − =− −− −

=−

( 2)2( 2) 1

( 2) 4

5

02 , which is also undefined.

• Another way to identify the asymptotes and zeros of a rational function is to look at a table of values generated on a graphing calculator. Follow the steps appropriate to your calculator model.

On a TI-83/84:

Step 1: Press [Y=]. Press [CLEAR] to delete any other functions stored on the screen.

Step 2: At Y1, use your keypad to enter values for the function. Use [X, T, θ, n] for x and [x2] for any exponents.

Step 3: To view a table of values for the function, press [2ND][GRAPH].

Step 4: Arrow up and down the list of ordered pairs to find the zeros and the domain values for which the function is undefined.

On a TI-Nspire:

Step 1: Press [home] to display the Home screen.

Step 2: Arrow down to the graphing icon, the second icon from the left, and press [enter].

Step 3: Enter the function to the right of “f1(x) =” and press [enter].

Step 4: To see a table of values, press [menu], then use the arrow keys to select 2: View, then 9: Show Table.

Step 5: Arrow up and down the list of ordered pairs to find the zeros and the domain values for which the function is undefined.


Example 1

Show that the function f xx

= −+

( ) 12

3 is a rational function.

1. Determine if the given function is a rational function in its current form.

In order for f(x) to be a rational function, it has to be a quotient of two polynomials.

The given function consists of a constant, 1, and a rational

expression, x

−+2

3; therefore, the function is not presented in the

form of a rational function.

2. If possible, rewrite the function so that it is a rational function.

A rational function must be a quotient of two polynomials.

Rewrite the constant, 1, as a quotient of two polynomials.

f xx

= −+

( ) 12

3 Original function

f xx

= −+

( )1

1

2

3 Rewrite the constant, 1, as a fraction.

f xx

x x=

++

−+

( )1( 3)

1( 3)

2

3Multiply by the common factor.

f xx

x x=

++

−+

( )( 3)

( 3)

2

3 Simplify.

f xx

x=

+ −+

( )3 2

3 Subtract the fractions.

f xx

x=

++

( )1

3Simplify.



3. Verify that the rewritten function is a rational function.

In order for f(x) to be a rational function, it must be a quotient of two polynomial functions.

The binomial x + 1 is a polynomial function.

The binomial x + 3 is a polynomial function.

4. Draw conclusions.

The original function can be rewritten as a rational function.

Therefore, the original function f xx

= −+

( ) 12

3 is a rational

function.

Example 2

Write a rational function, h(x), using the functions f(x) = 2x – 1 and g(x) = 4x2.

1. Write the definition of a rational function using f(x), g(x), and h(x).

There are two possible ways to write the rational function, h(x), in terms of f(x) and g(x).

One way is h xf x

g x=( )

( )

( ), where f(x) is the numerator and g(x) is the

denominator.The second way is h xg x

f x=( )

( )

( ), where the positions of

f(x) and g(x) have been swapped—that is, g(x) is the numerator and

f(x) is the denominator.


2. Write each possibility for the rational function, h(x), in terms of the given functions.

Use the definition of a rational function and the given polynomials to write each possibility.

First possibility:

h xf x

g x=( )

( )

( )Definition of a rational function, h(x)

h xx

x( )( )

=−

( )2 1

4 2 Substitute 2x – 1 for f(x) and 4x2 for g(x).

Second possibility:

h xg x

f x=( )

( )

( )Definition of a rational function, h(x)

h xx

x( )( )=

−( )

4

2 1

2

Substitute 4x2 for g(x) and 2x – 1 for f(x).

The two possibilities for the rational function h(x) are h xx

x=

−( )

2 1

4 2

and h xx

x=

−( )

4

2 1

2

.


3. Verify that both possible functions for h(x) are rational functions.

Simplify each function for h(x).

h xx

x=

−( )

2 1

4 2 First possible rational function

h xx

x x= −( )

2

4

1

42 2

Rewrite the fraction as a subtraction of two fractions.

h xx

x x= −( )

2

1

42 2 Simplify.

h xx x

= −( )1

2

1

4 2

By definition, this possibility is not a rational function because in its simplest form, it is not the quotient of two polynomials.

Look at the second possible rational function, h xx

x=

−( )

4

2 1

2

.

This possibility cannot be simplified any further.

By definition, this possibility is a rational function because in its simplest form, it is the quotient of two polynomials.

Therefore, the only rational function that can be written for h(x)

using the functions f(x) = 2x – 1 and g(x) = 4x2 is h xx

x=

−( )

4

2 1

2

.


Example 3

Write a rational function that has a zero of x = –1 and is undefined at x = 2.

1. Use the definition of a rational function to write the simplest general polynomials for the numerator and the denominator.

Recall the definition of a rational function is h xf x

g x=( )

( )

( ).

The simplest polynomials for the numerator and the denominator have the form f(x) = x + a and g(x) = x + b.

Therefore, h xf x

g x

x a

x b= =

++

( )( )

( ).

2. Determine the polynomial for the numerator.

Use the zero of the function to determine the numerator.

We are given that a zero of this function is at x = –1; therefore, f(–1) = 0.

f(x) = x + a Numerator of the rational function

(0) = (–1) + a Substitute 0 for f(x) and –1 for x.

a = 1 Add 1 to each side to solve for a.

Now that we know a = 1, substitute 1 for a in the numerator of the rational function.

f(x) = x + a Numerator of the rational function

f(x) = x + (1) Substitute 1 for a.

f(x) = x + 1 Simplify.

Therefore, f(x) = x + 1, and x + 1 is the polynomial for the numerator.


3. Determine the polynomial for the denominator.

The denominator is undefined for values that would result in a 0 in the denominator.

We are given that the function is undefined at x = 2; therefore, g(2) = 0.

g(x) = x + b Denominator of the rational function

(0) = (2) + b Substitute 0 for g(x) and 2 for x.

b = –2 Subtract 2 from each side to solve for b.

Now that we know b = –2, substitute –2 for b in the denominator of the rational function.

g(x) = x + b Denominator of the rational function

g(x) = x + (–2) Substitute –2 for b.

g(x) = x – 2 Simplify.

Therefore, g(x) = x – 2, and x – 2 is the polynomial for the denominator.

4. Write the rational function using the polynomials determined for the numerator and denominator.

Substitute the functions determined in steps 2 and 3 for the numerator and denominator of the definition of a rational function.

h xf x

g x=( )

( )

( )Definition of a rational function

h xx

x=

+−

( )( 1)

( 2)Substitute x + 1 for f(x) and x – 2 for g(x).

This rational function is already in simplest form; therefore, the

rational function, h xx

x=

+−

( )1

2, has a zero of x = –1 and is

undefined at x = 2.


Example 4

Identify the vertical asymptote(s) for the rational function f xx x

x x x=

+− + −

( )( 1)

( 2)( 3)( 4).

1. Identify the values of x for which the rational function is undefined.

The rational function is undefined for values that would result in a 0 in the denominator.

Therefore, any factors in the denominator (x – 2, x + 3, and x – 4) that are equal to 0 would result in a value of x that gives an undefined value of f(x).

Set each factor equal to 0 and solve the equation for x to determine each value that makes f(x) undefined.

x – 2 = 0 x + 3 = 0 x – 4 = 0

x = 2 x = –3 x = 4

The values of x = 2, x = –3, and x = 4 result in an undefined rational function.

2. List the vertical asymptotes for the rational function.

The vertical asymptotes are located at the values of x for which the function is undefined and are of the form x = a.

Therefore, the vertical asymptotes of the rational function,

f xx x

x x x=

+− + −

( )( 1)

( 2)( 3)( 4), are x = 2, x = –3, and x = 4.

U3-97

PRACTICE

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPSLesson 3: Graphing Rational Functions


For problems 1–3, write a rational function, f(x), in simplest form using the two given functions.

1. g(x) = –3x; h(x) = x + 3

2. g(x) = 7 – x; h(x) = 7 + 21x

3. g(x) = –1; h x xx

= −( )2

3

For problems 4–7, complete each problem involving rational functions.

4. Peyton buys a smartphone for $300. She signs a contract for wireless and voice services that will cost $45 per month. Write a rational function for the monthly cost of the smartphone and the data plan, C(m), over m months.

5. Write a rational function, f(n), for 3 times a number n less twice its reciprocal.

6. Risa and Madison are fiber-optic cable installers. Risa takes 1 hour less to install

a floor of a new office building than Madison does. The number of hours (y) it

takes them to install a floor of the office building when working together is given

by the equation x x y−

+ =1

1

1 1, where x is the number of hours Madison takes

when working alone and x – 1 is how long Risa takes when working alone. Show

that y can be written as a rational equation.

Practice 3.3.1: Creating Rational Equations in Two Variables

continued

U3-98

PRACTICE



7. Xavier bikes 100 kilometers in the outbound leg of the “Summit to Surf ” bike ride for t hours at a rate of r1 kilometers per hour. His time on the return trip is 2 hours longer than on the first leg of the ride at a speed of r2. If the speed is distance (100 km) divided by time, show that the average speed r for the round trip can be written as a rational equation. (Hint: Write a fraction for r1 and r2. What is the average of r1 and r2?)

For problems 8–10, write a rational function of the form f xg x

h x=( )

( )

( ) that has the

given vertical asymptote(s) and zero(s).

8. vertical asymptotes: x = 0, x = –1; zero: x = 1

9. vertical asymptote: x = –2; zeros: x = 4, x = 8

10. vertical asymptotes: x = 10, x = 20; zeros: x = 0.1, x = 0.01


Introduction

The graphs of rational functions can be sketched by knowing how to calculate horizontal and/or vertical asymptotes of the function and its zeros. The sketch of the graph can be improved by plotting points with x- and y-values that are at or near the limit of those domain and range values. Graphing calculators can be used to sketch the graph of a rational function, and can also provide the user with a table of x- and y-values from which these critical features of the graph can be read.

Key Concepts

• There are several key features of the graph of a rational function that are helpful when creating a sketch, including its zeros, horizontal and vertical asymptotes, domain, range, and extreme values.

• The range is the set of all outputs of a function; that is, the set of y-values that are valid for the function.

• Recall that the zeros of the rational function are the domain value(s) for which

f(x) = 0. If the rational function f(x) = 0 is represented by f xg x

h x=( )

( )

( ), then

g(x) = 0 and h(x) ≠ 0.

• If a rational function f(x) = 0 is represented by f xx a x b

x c x d=

− −− −

( )( )( )

( )( ), then

(x – a)(x – b) = 0 and (x – c)(x – d) ≠ 0. The zeros are represented by the points

(a, 0), (b, 0), etc., on the graph.

• Asymptotes are straight lines on the graph of a rational function that form the boundaries for parts of the graph. In some cases, the domain values of the asymptotes are the domain values for which the rational function is undefined. In other cases, the domain values of the asymptotes form the maximum and/or minimum value(s) of the rational function.

• Vertical asymptotes are straight lines of the form x = a. For example, the graph

of the rational function f xx

x=

−( )

4

42 has vertical asymptotes at x = ±2, which

are solutions to the equation x2 – 4 = 0. The function f(x) is undefined at these

domain values.

Lesson 3.3.2: Graphing Rational Functions


• Horizontal asymptotes are straight lines of the form y = b. For example, the

graph of the function g xx

x=

−−

( )4 7

2 has a horizontal asymptote at y = 4. As

seen in the following graph, g(x) approaches y = 4.

–10 –8 –6 –4 –2 2 4 6 8 10

x

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

y

0


• Slant asymptotes are asymptotes of a graph of the form y = ax + b, where a ≠ 0. A rational function that has a numerator with a degree that is exactly 1 more than the degree of the denominator will have a slant asymptote.

–10 –8 –6 –4 –2 2 4 6 8 10

x

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

y

0

• To find the slant asymptote, use polynomial long division. The result of dividing the numerator by the denominator is the equation of the slant asymptote.

• Rational functions may or may not have a vertical asymptote, but will always have either a horizontal or a slant asymptote. A rational function will never have both a horizontal asymptote and a slant asymptote.

• For the graph of a rational function, it can be useful to plot several points that represent values of x and y that are far away from zeros and asymptotes. Such a value is known as an extreme value—a point on the graph of a rational function that occurs at or near the endpoints of its domain or range. These so-called extreme values will ensure that the sketch will be accurate over the domain and range of the function.

• For example, consider the function f xx

x=

−( )

1. The domain of the function

is all numbers except for x = 1 since the function is undefined for this value

of x. For values of x very close to x = 1, the value of the function is large. For


x = 1.01, f(x) = 101, and for x = 0.99, f(x) = 99. But, for x = 100, f(x) = 1.1, and for

x = –100, f(x) = 0.99. Sketching the points (1.01, 101), (0.99, 99), (100, 1.1), and

(–100, 0.99) shows that f(x) has a horizontal asymptote at y = 1 and a vertical

asymptote at x = 1.

–10 –8 –6 –4 –2 2 4 6 8 10x

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

y

0

• An example of a function without a vertical asymptote is g xx

x=

+( )

6

12

because there are no values of x that will result in a denominator that equals 0.

However, g(x) does have horizontal asymptotes at x = ±1 . The extreme values

of g(x) occur at x = ±1: g =+

=(1)6(1)

1 132 and g − =

−− +

=−( 1)6( 1)

( 1) 132 . The

extreme values of x occur over the domain −∞ ∞( , ) . If x = 100, g(100) = 0.06;

if x = –100, g(–100) = –0.06. In other words, the graph of g(x) approaches 0 as

x approaches ±100.

• These ideas can be explored and extended using a graphing calculator. Refer to the previous sub-lesson for directions specific to your calculator model.


Example 1

Sketch the graph of the rational function f xx

x=

−+

( )2

2

2

2 . Include points on both sides of any asymptotes.

1. Find the value of f(0).

The point for which x = 0 is the y-intercepts of the graph, and has the form (0, f(0)). The function value f(0) may or may not be equal to 0.

=−

+= =(0)

2 (0)

(0) 2

2

21

2

2f

The point on the graph representing f(0) is (0, 1).

2. Find the value(s) of x for which f(x) = 0.

The values of x for which f(x) = 0 are the zeros of the function.

If f(x) = 0, x2 + 2 ≠ 0, so 2 – x2 = 0, which means that x = ± 2 .

Therefore, the zeros of f(x) are given by ( )2,0 and ( )− 2,0 .

3. Write the equation(s) for the vertical asymptote(s) of f(x), if any.

Vertical asymptotes of f(x), if any, occur at values of x which make the denominator equal 0.

In this case, setting the denominator equal to 0 and solving for x (x2 + 2 = 0) will not yield values of x that are within the domain of the function, which is −∞ ∞( , ) .

Therefore, there are no vertical asymptotes for f(x).



4. Write the equation(s) of the horizontal asymptote(s) of f(x), if any.

The horizontal asymptotes are of the form y = b. One way to find b is to solve the rational function for the domain variable, x.

f xx

x=

−+

( )2

2

2

2Original function

f x x

x=

−+

( )

1

2

2

2

2Rewrite the function as two fractions.

f(x) • (x2 + 2) = 1 • (2 – x2) Cross multiply.

x2 • f(x) + 2 • f(x) = 2 – x2 Simplify.

x2 • f(x) + x2 = 2 – 2 • f(x) Use subtraction and addition to group like terms (x2 and 2 • f(x)) on the same sides of the equation.

x2 • [f(x) + 1] = 2 • [1 – f(x)] Simplify.

[ ]=

• −+

2 1 ( )

( ) 12x

f x

f xDivide.

[ ]=

• −+

2 1 ( )

( ) 1x

f x

f xSolve for x by taking the square root of both sides.

This rational expression for x is undefined when the denominator f(x) + 1 = 0, so the horizontal asymptote is y = –1.

5. To prepare for sketching the graph of f(x), pick values for x that meet the conditions for f(x) = 0 from step 1, along with some values of x <− 2 , − < <2 2x , and <2 x. Then, calculate the value of f(x) for each value of x. List the calculations in a table.

Let x = –2, –1, 0, 1, 2, and 3 and solve for the corresponding f(x) value in the original function.

x –2 –1 0 1 2 3f(x) –0.33 0.33 1 0.33 –0.33 –0.64


6. List the points calculated in steps 1 and 2 for the y-intercepts and zeros of the function in a table.

x − 2 0 2f(x) 0 1 0

7. Use your results to sketch the graph of the function f xx

x=

−+

( )2

2

2

2 .

To sketch the graph, first plot the points listed in the tables in steps 5 and 6.

The conditions listed in steps 5 and 6 will ensure that all of the regions on the coordinate plane that contain parts of the rational function’s graph will be represented.

Use dotted lines to sketch the horizontal asymptote, y = –1.

Use a graphing calculator to check your work. Refer to the appropriate calculator directions for your calculator model as provided in the Key Concepts in the previous sub-lesson.

Your graph should resemble the following:

–10 –8 –6 –4 –2 2 4 6 8 10

x

20

18

16

14

12

10

8

6

4

2

0

–2

–4

–6

–8

–10

–12

y


Example 2

Sketch the graph of the rational function f x xx

= +( )1

, which represents the sum of a number and its reciprocal.


The point for which x = 0 is the y-intercept of the graph, and has the form (0, f(0)).

f = +(0) 01

0, which is undefined.

There is no point on the graph representing f(0).

Therefore, x = 0 is a vertical asymptote.



Rewrite the function as one fraction.

f x xx

= +( )1

Original function

f xx

x= +( )

1

1Write the term x as a fraction.

f xx

x

x

x= +( ) •

1

1Multiply

x

1 by

x

x (which is equivalent to 1).

f xx

x x= +( )

12

Multiply.

f xx

x=

+( )

12

Simplify.

If f(0) = 0, then x2 + 1 = 0 (assuming that x ≠ 0 from step 1).

The equation x2 + 1 = 0 has no solution on the interval (–∞, ∞), so this rational function has no zeros.


3. Find any other asymptote(s) by solving the function for x.

Rearrange and solve the function for x to reveal what values of f(x) are possible for the function.

This will also indicate any values that the function cannot have.

f xx

x=

+( )

12

Function from the previous step

f x x

x=

+( )

1

12

Rewrite the function as two fractions.

f(x) • (x) = 1 • (x2 + 1) Cross multiply.

x[f(x)] = x2 + 1 Simplify.

0 = x2 – x[f(x)] + 1Use subtraction to collect all like terms on one side of the equation.

Two values of f(x) that allow this quadratic equation to be solved are f(x) = ±2.

Values of f(x) ≥ 2 and f(x) ≤ –2 result in domain values (values of x) that are in the domain of the function.

However, values of f(x) such that –2 < f(x) < 2 cannot be achieved with the domain values of the function.

Notice that the function f xx

x=

+( )

12

has a numerator that is exactly

one degree higher than that of the denominator. This indicates a slant

asymptote.

To determine the equation of the slant asymptote, use polynomial long division to divide x2 + 1 by x.

)x x

x

x+1

1

2

2

The slant asymptote is the polynomial part of the answer, x, not the remainder.

Therefore, the slant asymptote is y = x.


4. Use the results of steps 1–3 to create a table of at least 6 points that can be used to sketch the graph.

Recall from step 1 that there is a vertical asymptote at x = 0.

There is also a gap in the range of function values between f(x) = 2 and f(x) = –2.

Points on either side of these range values should begin to suggest the shape of the graph.

Therefore, choose values of x that result in range values close to these function values and solve for f(x). Summarize your results in a table.

x –3 –2 –1 0 1 2 3f(x) –3.33 –2.5 –2 — 2 2.5 3.33

5. Use your results to sketch the graph of the function f x xx

= +( )1

.

To sketch the graph, first plot the points listed in the table in step 4.

Use a dotted line to sketch the vertical asymptote, x = 0, and the slant asymptote, y = x.


Your graph should resemble the following:

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10


Example 3

Sketch the graph of the rational function f xx

x=

−−

( )25

9

2

2 . Include points on both sides of any asymptotes.


The point for which x = 0 is the y-intercept of the graph, and has the form (0, f(0)).

The function value f(0) may or may not be equal to 0.

f xx

x=

−−

( )25

9

2

2 Original function

f =−−

(0)(0) 25

(0) 9

2

2 Substitute 0 for x.

f =−−

(0)0 25

0 9Simplify.

f =(0)25

9The value of f(0) is

25

9.

Therefore, the point on the graph representing f(0) is 0,25

9

.



If f(x) = 0, and the denominator (x2 – 9) cannot equal 0, then the numerator (x2 – 25) must equal 0.

Determine the values that make the numerator equal to 0.

x2 – 25 = 0 Numerator

x2 = 25 Add 25 to both sides.

x = 252 Take the square root of both sides.

x = ±5 Simplify.

The values of x that make the numerator equal 0 are 5 and –5.

Therefore, the zeros of f(x) are given by (5, 0) and (–5, 0).


3. Write the equation(s) for the vertical asymptote(s) of f(x), if any exist.

The values of x for the vertical asymptote(s) occur when the denominator is equal to 0.

Determine the values that make the denominator equal to 0.

x2 – 9 = 0 Denominatorx2 = 9 Add 9 to both sides.

x = 92 Take the square root of both sides.

x = ±3 Simplify.

The values of x that make the denominator equal 0 are 3 and –3.

Therefore, the vertical asymptotes are x = 3 and x = –3.

4. Write the equation(s) of the horizontal asymptote(s) of f(x), if any.

The horizontal asymptotes are of the form y = b.

One way to find b is to solve the rational function for the domain variable, x.

f xx

x=

−−

( )25

9

2

2 Original function

f x x

x=

−−

( )

1

25

9

2

2

Rewrite the function as two fractions.

f(x) • (x2 – 9) = 1 • (x2 – 25) Cross multiply.

x2 • f(x) – 9 • f(x) = x2 – 25 Simplify.

x2 • f(x) – x2 = 9 • f(x) – 25Use subtraction and addition to group like terms on the same sides of the equation.

x2 • [f(x) – 1] = 9 • f(x) – 25 Simplify.

xf x

f x

9 ( ) 25

( ) 12 =

• −−

Divide.

xf x

f x

9 ( ) 25

( ) 1=

• −−

Isolate x by taking the square root of both sides.

This rational expression for x is undefined when f(x) = 0, so the horizontal asymptote is y = 1.


5. To prepare for the sketch of the graph for f(x), pick values for x that meet the conditions x > 3, –3 < x < 3, and x < –3. Then, calculate the value of f(x) for each value of x. List the calculations in a table.

Let x = –5, –4, –2, –1, 0, 1, 2, 4, and 5. Substitute each value of x into the original equation and solve for f(x).

x –5 –4 –2 –1 0 1 2 4 5f(x) 0 –1.29 4.2 3 2.78 3 4.2 –1.29 0

6. Finally, list the points calculated in steps 1 and 2 for the y-intercepts and zeros of the function and in a table.

x –5 0 5f(x) 0 2.78 0

7. Use your results to sketch the graph of the function f xx

x=

−−

( )25

9

2

2 .

To sketch the graph, first plot the points listed in the tables in steps 5 and 6.

The conditions listed in steps 5 and 6 will ensure that all of the regions on the coordinate plane that contain parts of the rational function’s graph will be represented.

Use dotted lines to sketch the horizontal asymptote, x = 1, and the vertical asymptotes, y = –3 and y = 3.


–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

U3-112

PRACTICE



Practice 3.3.2: Graphing Rational Functions

For problems 1–4, sketch the graph of the given rational function on a coordinate plane. Include asymptotes, zeros, and extreme points, if any.

1. f xx

x x=

−+ −

( )9 16

3 4

2

2

2. g xx

x x=

++ +

( )9

2 4

2

2

3. f xx x

= −( )1 1

3 2

4. g xx x

x=

− −+

( )2 2

2

2

continued

U3-113

PRACTICE



For problems 5 and 6, name the horizontal, slant, and/or vertical asymptote(s) and the zero(s) of each graphed rational function.

5.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

6.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

continued

U3-114

PRACTICE



Use the following information to complete problems 7–10.

A distributor of vegetable oil uses cylindrical containers that have a volume of 0.5 m3. The distributor would like to find out what container radius would result in the smallest container surface area.

7. Use the formulas for the surface area of a cylinder, S(r, h) = 2πr2 + 2πrh, and the volume of a cylinder, V(r, h) = πr2h, to write a rational function for the surface area in terms of the radius r of the cylindrical containers.

8. Sketch a graph on a coordinate plane for all of the surface areas that are possible for different radii.

9. Describe how the surface area of the container changes as the radius of the container changes from less than 1 meter to greater than 1 meter.

10. What is the significance of the parts of the rational function graph that have domain values given by r ≤ 0?


Introduction

The zero(s) of a rational function can be identified in different ways, some of which involve manipulation of the terms in the function. Factoring, finding common denominators, and using the quadratic formula are methods that can be used in this search. A graph of the rational function can also be used to visually identify the location of zeros, or to indicate that they do not exist. A data table accompanying a graph can be used to approximate zeros of rational functions when the calculation techniques are too involved, or when those techniques increase the chances of making algebraic or arithmetic errors.

Key Concepts

• Recall that the zeros of a rational function are the domain value(s) for which

f(x) = 0. If the rational function f(x) = 0 is represented by f xg x

h x=( )

( )

( ), then

g(x) = 0 and h(x) ≠ 0.

• If a rational function f(x) = 0 is represented by f xx a x b

x c x d=

− −− −

( )( )( )

( )( ), then

(x – a)(x – b) = 0 and (x – c)(x – d) ≠ 0. The zeros are represented by the points

(a, 0), (b, 0), etc., on the graph.

• Recall that a factor is one of two or more numbers or expressions that when multiplied produce a given product.

• Factoring, or the process of separating an expression into two or more prime

factors, can sometimes be used to simplify the numerator of a rational function

so that the zero(s) can be identified. For example, the numerator of the rational

function f xx x

x=

− −−

( )6 7 5

9

2

can be factored to give f xx x

x=

+ −−

( )(2 1)(3 5)

9.

The zeros can be found by solving the equations 2x + 1 = 0 and 3x – 5 = 0 for x.

• Sometimes the factoring involves a monomial term. For example, the

numerator of the function g xx x

x=

++

( )2 2

5

3 2

2 can be factored by removing the

monomial 2x2 out of the numerator to yield g xx x

x=

++

( )2 ( 1)

5

2

2 . Then, to find

the zeros, solve the equations 2x2 = 0 and x + 1 = 0 for x.

Lesson 3.3.3: Finding the Zeros


• Recall that the quadratic formula can be used to solve some second-

degree polynomials, and also when a polynomial will not factor easily. For

a quadratic equation of the form ax2 + bx + c = 0, the value of x is given by

xb b ac

a=− ± − 4

2

2

. In the rational equation h xx x

x=

+ −+

( )2 2

2 3

2

, the quadratic

formula can be used to show that the value of x in the numerator, 2x2 + x – 2,

is x =− ± − −1 1 4(2)( 1)

4

2

, which is equal to − ±1

4

17

4. This means that the

zeros of 2x2 +x – 2 are x =− +1

4

17

4 and x =− −

1

4

17

4.

• It is also useful to know if all of the zeros of a rational function have

been found. This can be determined by looking at the degree of the

numerator’s polynomial function. The degree of a polynomial is the

highest power of the variable(s) in an expression. For example, in the

rational function f xx x x

x=

− − +−

( )3 3

4

3 2

2 , the degree of the numerator’s

polynomial is 3. Factoring the numerator shows that there are three zeros:

f xx x x

x

x x x

x

x x

x

x x x

x=

− − +−

=− − −

−=

− −−

=− + −

−( )

3 3

4

( 3) ( 3)

4

( 1)( 3)

4

( 1)( 1)( 3)

4

3 2

2

2

2

2

2 2

• The zeros are x = –1, 1, and 3.

• In other cases of third-degree numerator polynomials, only one zero can be identified. In these cases, the other two zeros are not real numbers. Zeros that are not real numbers occur in pairs, which can help you identify how many real zeros a numerator’s polynomial can have. The real zeros are solutions to an equation that can be graphed on a coordinate plane of real numbers. For example, a numerator’s polynomial of degree 4 can have 4 real zeros, 2 real zeros, or no real zeros. The polynomial still has 4 zeros, but not all of them are real and will not appear on the graph of the function.


• If calculations of zeros by factoring and/or using the quadratic formula prove to be difficult, then a graphing calculator or graphing utility software can be used. These tools provide a visual approach to identifying zeros and the values of the x- and y-coordinates of a rational function.

• To find the zeros of a rational function, follow the directions appropriate to your calculator model.

On a TI-83/84:


Step 2: At Y1, use your keypad to enter values for the function. Use [X, T, θ, n] for x and [x2] for any exponents. Press [GRAPH].

Step 3: To adjust the graph’s axes so the graph is easier to study, press [WINDOW], and change the axis settings for Xmin, Xmax, Ymin, and Ymax as needed.

Step 4: To find the zeros, press [2ND][GRAPH] to display a table of values.

Step 5: To find where the zeros are located, find the places where the value of Y1 changes from positive (+) to negative (–).

Step 6: To narrow down the zeros even more, press [TRACE]. Use the arrow keys to move the cursor along the two parts of the graph until the cursor crosses the x-axis and the Y1 values change from positive to negative.


On a TI-Nspire:




Step 4: To adjust the x- and y-axis scales on the window, press [menu] and select 4: Window and then 1: Window Settings. Enter each setting as needed. Tab to “OK” and press [enter].

Step 5: To see a table of values, press [menu] and scroll down to 2: View, then 9: Show Table.

Step 6: To find where the zeros are located, arrow up and down the table to find the places where the value of Y1 changes from positive (+) to negative (–).

Step 7: To narrow down the zeros even more, use the Trace tool. Press [ctrl][tab] to reactivate the graph, then [ctrl][t] to hide the table. Then, press [menu] and use the arrow keys to select 5: Trace, then 1: Graph Trace. Press [enter].

Step 8: Use the arrow keys to move the cursor along the two parts of the graph until the cursor crosses the x-axis. At that point, the symbol

zero appears, indicating the closest approximation of one of the solutions.


Example 1

Find the zero(s) of the rational function f xx x

x x=

+ −− +

( )3 2 8

4 3 2

2

2 .

1. Try to factor the quadratic function in the numerator before trying other methods of finding the zeros.

There are no common terms in the numerator, so the numerator may be a product of two binomials.

2. Determine if the numerator is the product of two binomials.

Reorder the numerator and denominator in descending order

according to degree. f xx x

x x=

+ −− +

( )3 2 8

4 3 2

2

2 can be rewritten as

f xx x

x x=− + +

− +( )

8 2 3

2 3 4

2

2 .

Next, list the factors of 3 and –8.

The factors of 3 are 1 and 3, and –1 and –3.

The factors of –8 are –1 and 8, –8 and 1, –2 and 4, and 2 and –4.

Next, try to find binomials with a product that results in a middle term of 2x.

For example, when multiplied, the binomials 3 – 8x and 1 + x yield a middle term of 3x – 8x or –5x. Continue checking binomial pairs until the correct pair is found.

The correct pair of factors is 3 – 4x and 1 + 2x, because when multiplied, the resulting middle term is –4x + 6x or 2x.



3. Once the correct pair of binomial factors is found, set each factor equal to 0 and solve.

The factors of the binomial are 3 – 4x and 1 + 2x.

Set each factor equal to 0 and solve for x.

3 – 4x = 0 Set the first factor equal to 0.

3 = 4x Add 4x to both sides.

x = 0.75 Solve for x.

1 + 2x = 0 Set the second factor equal to 0.

2x = –1 Subtract 1 from both sides.

x = –0.5 Solve for x.

Any zeros of the numerator will also be zeros of the rational function if they are not also zeros of the denominator. Check that the zeros of the numerator, found above, do not also make the denominator equal 0.

Substitute –0.5 and 0.75 for x and simplify.

2x2 – 3x + 4 2x2 – 3x + 4

= 2(–0.5)2 – 3(–0.5) + 4 = 2(0.75)2 – 3(0.75) + 4

= 6 = 2.875

Substituting each of the zeros into the denominator does not make

the denominator equal 0; therefore, the zeros of the rational

function f xx x

x x=

+ −− +

( )3 2 8

4 3 2

2

2 are x = 0.75 and x = –0.5.


Example 2

Find the zero(s) of the rational function g xx x

x=

− −+

( )3

1

2

2 .


There are no common terms in the numerator, so the numerator may be a product of two binomials.

2. Determine if the numerator is the product of the two binomials.

First, list the factors of 1 and –3, since the coefficient of the x2 term is 1.

The factors of 1 are 1 and 1, and –1 and –1.

The factors of –3 are –1 and 3, and 1 and –3.

Next, try to find binomials with a product that results in a middle term of –x.

There is no combination of binomials that when multiplied yield a product of –x.

3. Use the quadratic formula to find the solutions of the equation x2 – x – 3 = 0.

The numerator, x2 – x – 3, is of the form ax2 + bx + c.

Use the quadratic formula, xb b ac

a=− ± − 4

2

2

, to solve for x. Let a = 1, b = –1, and c = –3.

xb b ac

a=− ± − 4

2

2

Quadratic formula

x =− ± − − −( 1) ( 1) 4(1)( 3)

2(1)

2

Substitute 1 for a, –1 for b, and –3 for c.

x =− ±1 13

2Simplify.

x =− ±1

2

13

2

(continued)


Any zeros of the numerator will also be zeros of the rational function if they are not also zeros of the denominator. Check that the zeros of the numerator do not also make the denominator equal 0.

Substitute each of the zeros for x and simplify.

x2 + 1 x2 + 1

=− +

+

1 13

21

2

=− −

+

1 13

21

2

≈ 2.69 ≈ 6.30

Substituting each of the zeros into the denominator does not make

the denominator equal 0; therefore, the zeros of the rational

function g xx x

x=

− −+

( )3

1

2

2 are x =− +1 13

2 and x =

− −1 13

2.

Example 3

Find the zero(s) of the rational function h xx x x

x=

− +−

( )2

1

4 3

.


Each term of the numerator’s trinomial contains the common factor x, so the numerator can be factored.

x4 – 2x3 + x Original numerator

x(x3 – 2x2 + 1) Factor out x.

A factor of x indicates that one of the zeros of the function is x = 0.


2. Determine if there is an obvious zero for the factor of x3 – 2x2 + 1.

Sometimes a quick inspection of the other factor(s) in the numerator will reveal another zero.

It can be seen that a value of x = 1 results in 0, as shown.

x3 – 2x2 + 1 = 0 Set the second factor equal to 0.

(1)3 – 2(1)2 + 1 = 0 Substitute 1 for x.

1 – 2 + 1 = 0 Simplify.

0 = 0

Therefore, in this case, x = 1 is also a zero.

3. Look for other factors of the numerator by determining if other binomials can be found.

In this example, the zero x = 1 can be made into a factor of x – 1. Next, use polynomial long division to divide x3 – 2x2 + 1 by x – 1.

)x x x x

x x

x x

x x

x x

x

x

1 2 0 1

0

1

1

1

1

0

3 2

3 2

2

2

2

− − + +

−

− +

− +

− −

− +− +

The numerator can be rewritten as x(x – 1)(x2 – x – 1).


4. Use the quadratic formula to check any remaining second-degree factors for zeros.


a=− ± − 4

2

2

, to solve the factor

x2 – x – 1 for x. Let a = 1, b = –1, and c = –1.

xb b ac

a=− ± − 4

2

2

Quadratic formula

x =± − − −(1) ( 1) 4(1)( 1)

2(1)

2

Substitute 1 for a, –1 for b, and –1 for c.

x =±1 5

2Simplify.

Any zeros of the numerator will also be zeros of the rational function if they are not also zeros of the denominator. Check that the zeros of the numerator do not also make the denominator equal 0.

Substitute each of the zeros for x and simplify.

x – 1 x – 1

1 5

21=

+

−

1 5

21=

−

−

≈ 0.62 ≈ 1.62

Substituting each of the zeros into the denominator does not

make the denominator zero, therefore, the zeros of the rational

function h xx x x

x=

− +−

( )2

1

4 3

are x =+1 5

2, x =

−1 5

2, x = 0,

and x = 1.


Example 4

Use a graphing calculator to find the zeros of the rational function

f xx x x x

= − + − +( )2.3

0.75

5.4

0.5

1.5

0.8

4.4

0.21.14 3 2 .

1. Enter the function on a graphing calculator to determine the number of zeros and their approximate values.

Due to the complexity of the constants and the coefficients in this problem, it can best be solved using a graphing calculator.

On a TI-83/84:


Step 2: At Y1, use your keypad to enter values for the function, as given in the problem statement. Use [X, T, θ, n] for x and [x2] for the exponents.

Step 3: Press [GRAPH] to see if the function appears in the standard 10-by-10 window. If not, press [2ND][GRAPH] to display a table of values.

Step 4: Scroll up and down the list of values until a trend appears that suggests how the graph will behave at large values of x and y. The magnitude of the x and y values will suggest a better scale for the axes so that all of the functions’ solutions can be seen.

Step 5: Press [WINDOW] to change the values of Xmin, Xmax, Ymin, and Ymax. The Xscl and Yscl values can also be changed to increase or decrease the magnitude of the intervals on the axes. For this function, change the axis settings for Xmin, Xmax, Ymin, and Ymax to –20, 20, –200, and 200, respectively. These values can be changed again after the zeros are located for a clearer view of the zeros’ values.

Step 6: Press [2ND][GRAPH] to display the x- and y-values again.

(continued)


Step 7: To find where the zeros are located, find the places where the value of Y1 changes from positive (+) to negative (–), or vice versa. At x = 0, y is undefined. At x = 1, y = –26.76. Therefore, for this function, a zero occurs between x = 0 and x = 1. At x = 19, y = –0.054. At x = 20, y = 0.003. Therefore, another zero occurs between x = 19 and x = 20. For values of x < 0 and x > 3, the signs of the y-values are the same.

Step 8: To narrow down the zeros even more, press [TRACE]. Move the cursor to the locations on the x-axis suggested by Step 7. For the location between x = 0 and x = 1, the value of x = 0.26 gives a value of y = 0.825 and the value of x = 0.27 gives a value of y = –26.3107. Therefore, one of the zeros is approximately 0.26. For the location between x = 19 and x = 20, the value of x = 19.93 gives a value of y = –0.0004 and the value of x = 19.95 gives a value of y = 0.0006. Therefore, the other zero is approximately 19.94.

On a TI-Nspire:



Step 3: Enter the function to the right of “f1(x) =” and press [enter]. For specific values, refer to the function given in the problem statement.

Step 4: To adjust the x- and y-axis scales on the window, press [menu] and select 4: Window and then 1: Window Settings. Enter [–][20] at XMin, [20] at XMax, [–][200] at YMin, and [200] at YMax. Tab to “OK” and press [enter].

Step 5: To see a table of values, press [menu] and scroll down to 2: View, then 9: Show Table.

Step 6: To find where the zeros are located, arrow up and down the table to find the places where the value of Y1 changes from positive (+) to negative (–), or vice versa.

(continued)


Step 7: To narrow down the zeros even more, use the Trace tool. Press [ctrl][tab] to reactivate the graph, then [ctrl][t] to hide the table. Then, press [menu] and use the arrow keys to select 5: Trace, then 1: Graph Trace. Press [enter].

Step 8: Use the arrow keys to move the cursor along the two parts of the graph until the cursor crosses the x-axis. At that point, the symbol zero appears, indicating the closest approximation of one of the solutions, which gives an x-value of approximately 0.26 when y is approximately –1.1E–9 (or 0 for all practical purposes).

Step 9: Continuing the trace to the right, the symbol zero appears again when x is approximately 19.94 and y is approximately 6.29E–5 (or 0 for all practical purposes).

– 10 – 8 – 6 – 4 – 2 2 4 6 8 10

x

200

180

160

140

120

100

80

60

40

20

– 20

– 40

– 60

– 80

– 100

– 120

– 140

– 160

– 180

– 200

y

0


2. Summarize your findings about the zeros for the rational function based on the graphing calculator explorations.

The graph produced using a graphing calculator suggests that this fourth-degree rational function has two real zeros: one at x ≈ 0.26 and another at x ≈ 19.9.

The graphing calculator gave more precise estimates of the location of the zeros for values of y that were approximately equal to 0. Because the graph of this function intersects the x-axis in only two locations, it suggests that the other two zeros for this function are not real numbers.

U3-129

PRACTICE



For problems 1 and 2, how many real zeros exist for the rational function? List any real zeros.

1. f xx

x=

+−

( )1

1

3

2. g xx x

x x=

− ++ −

( )5 4

2

4 2

2

For problems 3–7, find the zero(s) of each rational function.

3. a xx x

x x=

− ++ −

( )( 5)(2 7)

( 7)(2 5)

4. b xx x x

x x x=

+ ++ −

( )( 1) ( 2)

( 1)( 2)

2 2 2

5. c xx x

x=

− −−

( )6

2

2

6. d xx x x

x=

− − ++

( )2 19 20

5

3 2

7. e xx

x=

+−

( )3.456 7.98

3.04 0.91

4

2

Use the following information to complete problems 8–10.

The time required for the Home Town Hikers to make a 30-mile hike

depends on the total distance covered and the speed at which each hiker

travels. The rational function H(r) gives the total time for the trip in miles

per hour if the hikers move at two different speeds: H rr r

=−

++

( )20

1

10

2.

8. What is the zero of the function?

9. How does the zero relate to the real-world conditions of the problem?

10. What would be a reasonable domain for r in order for the rate and time to have magnitudes that make sense in this situation?

Practice 3.3.3: Finding the Zeros


Introduction

A system of rational equations or functions consists of two or more equations or

functions of the form f xg x

h x=( )

( )

( ). A solution of a system of rational functions exists

at domain values for which f(x) = g(x) = h(x). It is also possible for a system of rational

equations or functions to have no solutions or an infinite number of solutions. The

solution(s) to a system of rational equations or functions only exist at values of x

for which the equations or functions in the system are defined. The solution(s) of a

rational equation or function can be calculated exactly using algebraic methods such

as by factoring, by using the quadratic formula, and by employing similar techniques,

or they can be approximated graphically using technology.

Key Concepts

• The solution(s) to a system of rational functions f(x) and g(x) consist of

the value(s) of x that are in the domains of the functions and that result

in function values for which f(x) = g(x). For example, if f xx

x=

+( )

1 and

g xx

x=

−+

( )2

3, the solution for the system of f(x) and g(x) is (–0.5, –1). The

domain of both functions includes x = –0.5.

• On a graph, the solutions are the points at which the equations intersect.

• A system of rational equations for which no solutions exist can best be understood by looking at a graph of the system. The following graph is for the

system f xx

x=

+( )

2 and g x

x

x=

−( )

2.

Lesson 3.3.4: Solving a System of Rational Equations


–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

• As the graph shows, there are no points of intersection of the two functions, which is the indicator that a solution does not exist.

• A system of rational equations for which an infinite number of solutions exists can be visualized by looking at a table of values for the system. The following

table is for the system f xx

x=− −− +

( )2 8

2 10 and g x

x

x x=

−− +

( )16

9 20

2

2 .

x f(x) g(x)6 10 107 5.5 5.58 4 49 3.25 3.25

10 2.8 2.811 2.5 2.5

• As the table shows, for each value of x, f(x) = g(x).

• When a system is too complicated to be solved algebraically, graphing technology can show the solutions and provide an approximate value for them. Use the graphing calculator directions appropriate to your model to find the solutions of rational functions. Note: The precision with which the cursor is moved to the intersection point can also affect the results.


On a TI-83/84:


Step 2: At Y1, use your keypad to enter values for f(x). (Refer to the problem statement for the values of the functions.) Use [X, T, θ, n] for x and [x2] for any exponents.

Step 3: At Y2, use your keypad to enter values for g(x). Press [GRAPH].

Step 4: Press [WINDOW] to change the viewing window.

Step 5: To approximate the solutions for the system, press [TRACE]. Use the arrow keys to move the cursor over each of the two intersection points. The values for x and y will be displayed.

On a TI-Nspire:



Step 3: Enter the function f(x) to the right of “f1(x) =”. (Refer to the problem statement for the values of the functions.) Press the down arrow to bring up “f2(x) =”.

Step 4: Enter the function g(x) to the right of “f2(x) =” and press [enter].

Step 5: To adjust the x- and y-axis scales on the window, press [menu] and select 4: Window and then 1: Window Settings. Enter values for XMin, XMax, YMin, and YMax. Tab to “OK” and press [enter].

Step 6: To approximate the solutions, press the [menu] key, and select 5: Trace. Arrow over to 1: Graph Trace. Press [enter].

Step 7: Use the arrow keys to move the cursor along the graph to the intersection point of the two graphs. The symbol zero will appear at the closest approximation for each solution.


Example 1

Find the solution(s) to the rational functions f xx x

x=

−+

( )( 1)

2 and g x

x

x=

+( )

1.

1. Set f(x) = g(x).

f(x) must equal g(x) for a value of the domain variable, x, to result in a solution of a system of rational equations.

f(x) = g(x) Set the functions equal to each other.

x x

x

x

x

−+

=

+

( 1)

2 1Substitute

x x

x

−+

( 1)

2 for f(x) and

x

x +1 for g(x).

The expression for f(x) = g(x) is x x

x

x

x

−+

=+

( 1)

2 1.


Use any algebraic techniques that will reduce the problem to a simpler one that can be solved for the domain variable.

To simplify this problem, begin by cross multiplying.

x x

x

x

x

−+

=+

( 1)

2 1Equation from the previous step

x(x – 1)(x + 1) = x(x + 2) Cross multiply.

x2 – 1 = x + 2 Divide each side by x and simplify.

x2 – x – 1 = 2 Subtract x from both sides.

x2 – x – 1 – 2 = 0 Subtract 2 from both sides.

x2 – x – 3 = 0 Simplify.

The problem has been reduced to a quadratic equation: x2 – x – 3 = 0.



3. Use the quadratic formula to solve the quadratic equation.


a=− ± − 4

2

2

, to solve for x. Let a = 1, b = –1, and c = –3.

xb b ac

a=− ± − 4

2

2

Quadratic formula

x =− ± − − −( 1) ( 1) 4(1)( 3)

2(1)

2 Substitute 1 for a, –1 for b, and –3 for c.

x =− ±1 13

2Simplify.

The zeros of x x

x

x

x

−+

=+

( 1)

2 1 are x =

− +1 13

2 and x =

− −1 13

2.

Since these radical terms would be difficult to use, they can be converted to decimal approximations: x ≈ 0.5 ± 1.8, which simplifies to x ≈ 2.3 or –1.3.

4. Determine any additional solutions.

It is possible to miss solutions to rational functions when dividing both sides of an equation by a term or terms.

To find any additional solutions, find the zeros of the term(s).

When simplifying f(x) = g(x), we divided both sides of the equation by x.

The zero of x is 0.


5. Substitute the x-values into the original equations to verify that they result in the same function values for f(x) and g(x).

The values of f(x) and g(x) will not be exactly equal since approximations for x are used.

Let x = 2.3.

f xx x

x=

−+

( )( 1)

2g x

x

x=

+( )

1

f[ ]

=−

+(2.3)

(2.3) (2.3) 1

(2.3) 2g =

+(2.3)

(2.3)

(2.3) 1 f(2.3) ≈ 0.7 g(2.3) ≈ 0.7

Let x = –1.3.

f xx x

x=

−+

( )( 1)

2g x

x

x=

+( )

1

f[ ]

− =− − −

− +( 1.3)

( 1.3) ( 1.3) 1

( 1.3) 2g − =

−− +

( 1.3)( 1.3)

( 1.3) 1 f(–1.3) ≈ 4.3 g(–1.3) ≈ 4.3

Let x = 0.

f xx x

x=

−+

( )( 1)

2g x

x

x=

+( )

1[ ]

=−

+(0)

(0) (0) 1

(0) 2f =

+(0)

(0)

(0) 1g

f(0) = 0 g(0) = 0The solutions to the system of f(x) and g(x) are (0, 0) and approximately {(2.3, 0.7), (–1.3, 4.3)}.


Example 2

Show that the rational functions f xx

x x=

++

( )1

( 2) and g x

x

x x=

−−

( )2

( 1) have no

common solution.

1. Set f(x) = g(x).

f(x) must equal g(x) for a value of the domain variable, x, to result in a solution of a system of rational equations.


x

x x

x

x x

++

=

−−

1

( 2)

2

( 1)Substitute

x

x x

++

1

( 2) for f(x) and

x

x x

−−

2

( 1)

for g(x).

The expression for f(x) = g(x) is x

x x

x

x x

++

=−−

1

( 2)

2

( 1).


Use any algebraic techniques in order to reduce the problem to a simpler one that can be solved for the domain variable.

To simplify this problem, begin by cross multiplying.

x

x x

x

x x

++

=−−

1

( 2)

2

( 1)Equation found in previous step

x(x – 1)(x + 1) = x(x + 2)(x – 2) Cross multiply.

(x – 1)(x + 1) = (x + 2)(x – 2) Divide both sides by x.

x2 – 1 = (x + 2)(x – 2) Distribute.

x2 – 1 = x2 – 4 Distribute.

–1 ≠ –4 Subtract x2 from both sides.

Note that –1 does not equal –4; therefore, any value of x chosen will not result in a true statement based on the two functions.


3. Verify your answer with a graphing calculator.

This step will give visual evidence that the graphs of the two functions do not intersect.

On a TI-83/84:


Step 2: At Y1, use your keypad to enter values for f(x). (Refer to the problem statement for the values of the functions.) Use [X, T, θ, n] for x and [x2] for any exponents.

Step 3: At Y2, use your keypad to enter values for g(x). Press [GRAPH].

Step 4: Press [WINDOW] to change the viewing window.

Step 5: To approximate the solutions for the system, press [TRACE]. Use the arrow keys to move the cursor over each of the two intersection points. The values for x and y will be displayed.

On a TI-Nspire:



Step 3: Enter the function f(x) to the right of “f1(x) =”. (Refer to the problem statement for the values of the functions.) Press the down arrow to bring up “f2(x) =”.

Step 4: Enter the function g(x) to the right of “f2(x) =” and press [enter].

Step 5: To adjust the x- and y-axis scales on the window, press [menu] and select 4: Window and then 1: Window Settings. Enter values for XMin, XMax, YMin, and YMax. Tab to “OK” and press [enter].

Step 6: To approximate the solutions, press the [menu] key, and select 5: Trace. Arrow over to 1: Graph Trace. Press [enter].

Step 7: Use the arrow keys to move the cursor along the graph to the intersection point of the two graphs. The symbol zero will appear at the closest approximation for each solution.

(continued)


Both calculators will yield a graph similar to the one below. The graphed functions do not intersect, and therefore have no real solutions.

–10 –8 –6 –4 –2 2 4 6 8 10x

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

y

0


Example 3

Show that the functions f xx x

x=

− −+

( )2

2

2

and g xx x x

x x=

+ − −+ +

( )2 5 6

5 6

3 2

2 have an infinite number of solutions.

1. Show that g(x) can be reduced to f(x), or that a common denominator can be found for f(x) and g(x).

The condition that f(x) and g(x) have an infinite number of solutions means that the functions are identical; therefore, any value of x will result in f(x) = g(x).

Begin by determining a binomial that can be multiplied by the denominator of f(x) in order to result in the denominator of g(x).

Notice that the denominator of g(x), x2 + 5x + 6, can be factored into x + 3 and x + 2.

Since the denominator of f(x) is x + 2, this suggests that x + 3 might be the factor that will make the two functions equal.

In order for the function value of f(x) to stay the same, the numerator and denominator have to be multiplied by the same value.

f xx x

x=

− −+

( )2

2

2

Original function for f(x)

f xx x x

x x=

− − ++ +

( )( 2)( 3)

( 2)( 3)

2Multiply the numerator and denominator by x + 3.

f xx x x

x x=

+ − −+ +

( )2 5 6

5 6

3 2

2Simplify both the numerator and denominator.

The function f xx x

x=

− −+

( )2

2

2

can be rewritten as

f xx x x

x x=

+ − −+ +

( )2 5 6

5 6

3 2

2 .

2. Compare the rewritten form of f(x) with g(x).

The rewritten form of f(x), x x x

x x

+ − −+ +

2 5 6

5 6

3 2

2 , is identical to g(x), x x x

x x

+ − −+ +

2 5 6

5 6

3 2

2 ; therefore, f(x) = g(x).


3. Verify that any value of x yields the same value for f(x) and g(x).

Any value of x selected should satisfy the condition that f(x) = g(x).

For instance, let x = 1. Substitute this value into the original expression for f(x).

f xx x

x=

− −+

( )2

2

2

Original function

f =− −+

(1)(1) (1) 2

(1) 2

2

Substitute 1 for x.

f =− −+

(1)1 1 2

1 2Simplify.

f =−(1)2

3

Now let x = 1 for every x-value in g(x) and simplify to see if the same result is produced.

g xx x x

x x=

+ − −+ +

( )2 5 6

5 6

3 2

2Original function

g =+ − −

+ +(1)

(1) 2(1) 5(1) 6

(1) 5(1) 6

3 2

2Substitute 1 for x.

g =− −

(1)3 5 6

12Simplify.

g =−(1)8

12

g =−(1)2

3

The value of f(1) is equal to g(1). We could substitute any value of x into both of these equations and arrive at the same answer for both functions. Therefore, this proves f(x) = g(x) for all values chosen for x.


Example 4

First, show that x = –1 is a solution to the system f xx

x=

+−

( )1

1

3

2 and g xx x

x x=

− −+ +

( )2

4 3

2

2 .

Then, state two different reasons why x = –1 cannot be used as a solution.

1. Evaluate f(–1) and g(–1).

Let x = –1.

f xx

x=

+−

( )1

1

3

2Given function for f(x)

f − =− +− −

( 1)( 1) 1

( 1) 1

3

2Substitute –1 for x.

f − =− +−

( 1)1 1

1 1Simplify.

f − =( 1)0

0

The result is undefined for f(–1).

g xx x

x x=

− −+ +

( )2

4 3

2

2Given function for g(x)

g( 1)( 1) ( 1) 2

( 1) 4( 1) 3

2

2− =− − − −− + − +


g(–1)1 1 2

1 4 3=

+ −− +

Simplify.

g(–1)0

0=

The result is undefined for g(–1).

When x = –1, the result for both functions is the same: undefined.

2. Explain why the results of step 1 do not represent a solution of the given functions.

Although the results are the same for both functions, the value of –1 results in an undefined quantity; therefore, x = –1 cannot be a solution.

The problem statement asks for two different reasons why x = –1 cannot be used as a solution. To find another reason, start by setting the two functions equal to each other.


3. Set f(x) = g(x).

This is the condition for a value of the domain (x) to result in a solution of a system of rational equations.


x

x

x x

x x

+−

=

− −+ +

1

1

2

4 3

3

2

2

2

Substitute x

x

+−

1

1

3

2 for f(x) and x x

x x

− −+ +

2

4 3

2

2 for g(x).

The expression for f(x) = g(x) is x

x

x x

x x

+−

=− −+ +

1

1

2

4 3

3

2

2

2 .


Use any algebraic techniques in order to reduce the problem to a simpler one that can be solved for the domain variable.

To simplify this problem, begin by factoring the numerators and denominators.

x

x

x x

x x

+−

=− −+ +

1

1

2

4 3

3

2

2

2

Equation found in the previous step

x x x

x x

x x

x x

+ − −− +

=− +− +

( 1)( 1)

( 1)( 1)

( 2)( 1)

( 1)( 1)

2Factor each numerator and denominator.

x x x

x x

x x

x x

+ − −− +

=− +− +

( 1) ( 1)

( 1) ( 1)

( 2) ( 1)

( 1) ( 1)

2

Cancel like factors.

x x

x

x

x

− −−

=−+

1

1

2

3

2

Simplify.


5. Explain the results of step 4.

The factor x + 1 representing the proposed solution x = –1 cancels out, which means it cannot be used as a solution for the system.

6. Check the results of steps 1–5 by evaluating the equation found in step 4 for x = –1.

Let x = –1.

x x

x

x

x

− −−

=−+

1

1

2

3

2

Equation found in step 3 for f(x) = g(x)

− − − −− −

=− −− +

( 1) ( 1) 1

( 1) 1

( 1) 2

( 1) 3

2


+ −− −

=− −− +

1 1 1

1 1

1 2

1 3Simplify.

−−

≠−−

1

2

3

2

Note that this is not a true statement; therefore, the domain value x = –1 is not a solution to the functions.

U3-144

PRACTICE



For problems 1–7, find the solution(s), if any, to each system of rational functions.

1. f xx

x=

−( )

12 ; g x

x

x=

−( )

1

3

2. f xx

x=

−( )

2

2; g x

x

x=

−( )

6

6

3. f xx

x=

−( )

2 13 ; g x

x x= −( )

2 12 3

4. f xx x x

x=

+ −+

( )( 2)( 5)

6; g x

x

x=

++

( )6( 2)

6

5. f xx

x=

−( )

1

2

2 ; g xx

x=

−( )

12

2

6. f xx

x=

−+

( )7 4

4 72 ; g xx

x=

−−

( )4 7

7 42

7. f xx x

x=

−−

( )( 1)

2

2

; g xx x

x=

−−

( )3 3

6 3

2 3

Practice 3.3.4: Solving a System of Rational Equations

continued

U3-145

PRACTICE



For problems 8–10, use the given information to find the solution(s) of the described systems of rational functions, if any solutions exist.

8. The school sports association paid $1,250 for a popcorn machine that costs $55 per month to operate. The association also purchased a hot chocolate drink machine for $725 that costs $80 per month to operate. Write rational functions for the total monthly cost of operating the popcorn machine P(n) and the hot chocolate machine H(n), including the original cost of each machine. Assuming a constant rate of use for each machine, when will the total monthly costs of operating the two machines be the same?

9. Gaby’s Lawn Service uses a herd of goats to “mow” a pasture of a specific size.

The number of hours, t, it takes 2 goats to mow the pasture is given by the

rational function A tt t

t=

−−

( )( 1)

2 1. If a third goat is used, the time can be given

by the function B tt t t

t=

− −−

( )( 1)( 2)

3 3. At what value of t do the 2 goats take the

same amount of time to finish the pasture as the 3 goats?

10. Two packages are wrapped for a birthday party. The ratio of the surface area

to the volume of each package is the same. The rational functions representing

the area-to-volume ratios for two different sizes of packages are represented by

the functions A xx

x x=

+− −

( )4( 5)

3( 2)( 4) and B x

x

x x=

+− −

( )5( 4)

4( 2)( 4). The constants

and binomial terms represent the dimensions of the packages in inches. If the

packages are rectangular solids, what are their dimensions when the ratios are

the same?

Lesson 4: Graphing Radical Functions





MCC9–12.F.IF.4★

MCC9–12.F.IF.5

MCC9–12.F.IF.7b

Essential Questions

1. What are the characteristics of the graph of a radical function and what is the significance of each?

2. What does an extraneous solution to a radical function mean graphically?

3. When does the domain of a radical equation not exist?

WORDS TO KNOW

cube root function a function that contains the cube root of a variable. The general form is y a x h k= − +( )3 , where a, h, and k are real numbers.

parent function a function with a simple algebraic rule that represents a family of functions. The graphs of the functions in the family have the same general shape as the parent function.

radical function a function with the independent variable under a root. The general form is f x a x h kn= − +( ) ( ) , where n is a positive integer root and a, h, and k are real numbers.

square root function a function that contains a square root of a variable

x-intercept the point at which a graphed function crosses the x-axis; written in the form (x, 0)

y-intercept the point at which a graphed function crosses the y-axis; written in the form (0, y)

U3-147Lesson 4: Graphing Radical Functions


• Khan Academy. “Graphing Radical Functions.”


This 10-minute video shows an example of setting up a table of values for a radical function, calculating the domain and range, and then plotting and graphing the function. An interactive script allows users to replay different parts of the video.

• Math Planet. “The Graph of a Radical Function.”


This site provides a summary of the graph of a radical function and includes a comparison of various radical functions, as well as an instructional video.

• Purplemath.com. “Graphing Radical Functions.”


This site describes how to graph different types of radical functions using a table of values.


Lesson 3.4.1: Creating Radical Equations in Two VariablesIntroduction

Radical equations and functions are often used to model various real-world scenarios. For example, a radical equation can be used to determine the distance seen from tall buildings based on the height of the viewing point. The equations of these scenarios are not always given, but can be created based on graphs or verbal descriptions.

Key Concepts

• A radical function is a function with the independent variable under a root. The general form is f x a x h kn( ) ( )= − + , where n is a positive integer root and a, h, and k are real numbers. The most common radical functions are the square root function, f x x=( ) , and the cube root function, g x x=( ) 3 .

• A square root function is a function that contains a square root of a variable.

• A cube root function is a function that contains the cube root of a variable. The general form is y a x h k= − +( )3 , where a, h, and k are real numbers.

• f x x=( ) and g x x=( ) 3 are parent functions, meaning f x x=( ) is the simplest square root function and g x x=( ) 3 is the simplest cube root function. Each parent function preserves the definition and shape of all other square root and cube root functions.

Understanding the Shape of a Square Root Function

• The typical graph of a radical function is one that curves slightly and goes on to either positive infinity or negative infinity for all real x-values.

• For the square root function f x x=( ) , the graph starts at the origin and curves out toward positive x-values and goes on to infinity.


• Notice how the graph of f x x=( ) angles upward as the values of x increase.

Parent Square Root Function, =( )f x x

2 4 6 8 10 12 14 16 18

12

10

8

6

4

2

0

f (x) = x

y

x

• A negative sign in front of the radical symbol, such as g x x=−( ) , causes the function to “flip” or be reflected over the x-axis and go on to positive infinity. On the other hand, a negative sign under the radical, such as h x x= −( ) , causes the function to flip over the y-axis and go on to negative infinity.

6

4

2

–2

–4

–6

2 4 6 80

g(x) = x

y

x

– 8 – 6 – 4 – 2

x

8

6

4

2

– 2

– 4

y

0

h(x) = x


• A radical function will start at a different point on the x-axis when numbers are added or subtracted under the radical sign. For instance, j x x= +( ) 1 would resemble the graph of f x x=( ) , but it would start 1 unit to the left. Thus, the first point would be the endpoint (–1, 0) instead of (0, 0). Subtracting 1 from x k x x( )= −e.g., ( ) 1 moves the graph 1 unit to the right, with the endpoint at (1, 0).

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

j(x) = x +1

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

k(x) = x – 1

• Also, if numbers are added to or subtracted from the term containing the radical sign, the graph shifts up and down along the y-axis. So, m x x= +( ) 4 resembles the graph of f x x=( ) , but it has been shifted 4 units up so that it has an endpoint of (0, 4) instead of (0, 0). Similarly, n x x= −( ) 4 resembles the graph of f x x=( ) , but it has been shifted 4 units down and thus has an endpoint of (0, –4).

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

m(x) = x + 4

– 2 2 4 6 8 10

x

6

4

2

– 2

– 4

– 6

y

0

n(x) = x – 4


• If the term containing the radical sign is multiplied by a factor, the graph of

the function will be vertically stretched or compressed by that same factor.

For instance, the graph of p x x=( ) 2 resembles the graph of f x x=( ) , but

it has been stretched vertically by a factor of 2. The graph of q x x=( )1

2 also

resembles the graph of f x x=( ) , but it has been compressed vertically by a

factor of 1

2.

–2 2 4 6 8 10x

10

8

6

4

2

–2

y

0

p(x) = 2 x

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

q(x) =12

x

• If x is multiplied by a factor of b, the graph of the function will be stretched or

compressed horizontally by that same factor. For instance, r x x=( ) 2 resembles

the graph of f x x=( ) , but it has been compressed horizontally by a factor of 1

2. The graph of s x x=( )

1

2 also resembles f x x=( ) ; however, it has been

stretched horizontally by a factor of 2.

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

r (x) = 2x

– 2 2 4 6 8 10

x

10

8

6

4

2

– 2

y

0

s(x) =12

x


• When determining the equation of a square root function using a graph, note where the graph starts and then determine in which direction the curve travels (toward positive or negative infinity for x-values).

• Also note whether the graph curves above or below the x-axis.

• The following table is a useful guide for interpreting various square root functions and understanding the graph of each. The table describes how the graph of each given square root function differs from that of the parent function, f x x=( ) .

Function Relation to the graph of the parent function, =( )f x x

f x x=−( ) f x x=( ) is reflected across the x-axis.

f x x= −( ) f x x=( ) is reflected across the y-axis.

f x x k= +( ) f x x=( ) is shifted k units to the left.

f x x k= −( ) f x x=( ) is shifted k units to the right.

f x x k= +( ) f x x=( ) is shifted k units up.

f x x k= −( ) f x x=( ) is shifted k units down.

f x a x=( ) f x x=( ) is stretched vertically by a factor of a.

f xa

x=( )1

f x x=( ) is compressed vertically by a factor of 1

a.

f x ax=( ) f x x=( ) is compressed horizontally by a factor of 1

a.

f xa

x=( )1

f x x=( ) is stretched horizontally by a factor of a.

Understanding the Shape of a Cube Root Function

• The typical graph of a cube root function, such as f x x=( ) 3 , is similar to a flat letter “S” that lies on its side.

• This cube root function travels from negative infinity to positive infinity, with the center at (0, 0) and points (1, 1) and (–1, –1).


Parent Cube Root Function, =( ) 3f x x

4

y

2

–2

–4

–10 –8 –6 –4 –2 2 4 6 8 100

x

• When a cube root function has a negative in front of the cube root, the graph is reflected over the x-axis.

• Notice that in the following graph of f x x=−( ) 3 , the graph still goes through the origin (0, 0) as well as the points (1, –1) and (–1, 1).

4

2

–2

–4

–10 –8 –6 –4 –2 2 4 6 8 100

f (x) = x3

y

x

• As with square root functions, cube root functions can be shifted along the x-axis by adding and subtracting numbers underneath the cube root.


• The graph of f x x= +( ) 43 resembles the graph of f x x=( ) 3 , but it has been shifted to the left 4 units. Its center is at (–4, 0), not (0, 0).

• If the number underneath the radical were subtracted, the graph would shift to the right.

4

2

–2

–4

–10 –8 –6 –4 –2 2 4 6 8 100

f (x) = x + 43

y

x

• If values are added to or subtracted from the term containing the cube root, the graph shifts up or down along the y-axis.

• The graph of f x x= +( ) 43 resembles the graph of f x x=( ) 3 , but it has been shifted up 4 units. Its center is at (0, 4) instead of (0, 0).

8

6

4

2

–2

–10 –8 –6 –4 –2 2 4 6 8 100

f (x) = x + 43

y

x


• Subtracting a number from the term containing the radical results in a graph that resembles the graph of f x x=( ) 3 , but it has been shifted down along the y-axis.

Function Relation to the graph of the parent function, =( ) 3f x x

f x x=−( ) 3 f x x=( ) 3 is reflected across the x-axis.

f x x= −( ) 3 f x x=( ) 3 is reflected across the y-axis.

f x x k= +( ) 3 f x x=( ) 3 is shifted k units to the left.

f x x k= −( ) 3 f x x=( ) 3 is shifted k units to the right.

f x x k= +( ) 3 f x x=( ) 3 is shifted k units up.

f x x k= −( ) 3 f x x=( ) 3 is shifted k units down.

f x a x=( ) 3 f x x=( ) 3 is stretched vertically by a factor of a.

f xa

x=( )1

3 f x x=( ) 3 is compressed vertically by a factor of 1

a.

f x ax=( ) 3 f x x=( ) 3 is compressed horizontally by a factor of 1

a.

f xa

x=( )1

3 f x x=( ) 3 is stretched horizontally by a factor of a.



Create the equation of a function in two variables that represents the graphed function.

6

4

2

–2

–4

–4 –2 2 4 6 8 10 12 140

y

x

1. Note the general shape of the curve and determine the type of function shown.

By looking at the graph, we can see that the function is a curved line and not a sideways letter “S.” Thus, it is a square root function.

2. Note the endpoint of the graph.

The endpoint of the graph is at (3, 0), which means that the parent function has been shifted along the x-axis 3 units to the right of the origin.

3. Note the general direction of the curve.

Since the curve rises from its endpoint, we know the square root is multiplied by a positive number. Since the curve flows right from its endpoint, we know x is also multiplied by a positive number.

4. Create an equation using the information derived from the graph.

The parent function of a square root function is f x x=( ) .

The graphed function is shifted 3 units to the right of the parent function; therefore, 3 is subtracted from x.

Based on this information, we can conclude that the radical equation represented in the graph is g x x( ) 3= − .


Example 2

Create the equation of a function in two variables that represents the graphed function.

8

6

4

2

–2

–6 –4 –2 2 4 6 8 100

y

x

1. Note the general shape of the curve and determine the type of function shown.

By looking at the graph, we can see that the function is a curved line and not a sideways letter “S.” Thus, it is a square root function.

2. Note the endpoint of the graph.

The endpoint of the graph is at (–5, 1), which means that the parent function has been shifted along the x-axis 5 units to the left and up 1 unit.

3. Note the general direction of the curve.

Since the curve rises from its endpoint, we know the square root is multiplied by a positive number. Since the curve flows right from its endpoint, we know x is also multiplied by a positive number.


Example 3

Determine the equation of a cube root function that is shifted 2 units to the left along the x-axis.

1. Identify the type of function to be created.

As stated, the type of function to be created is a cube root function.

The parent function for cube roots is f x x=( ) 3 .

2. Identify any shifts in the graph.

We are given that the function is shifted 2 units to the left along the x-axis. No other shifts are described.

Because the function is shifting along the x-axis, there is only a horizontal shift, not a vertical shift.

3. Use the known information to write the equation of the function.

The parent function of a cube root function is f x x=( ) 3 .

The function described is shifted 2 units to the left of the parent function; therefore, 2 is added to x.

Based on this information, we can conclude that the radical equation represented by the verbal description is g x x( ) 23= + .

4. Create an equation using the information derived from the graph.

The parent function of a square root function is f x x=( ) .

The graphed function is shifted 5 units to the left of the parent function; therefore, 5 is added to x, or g x x= +( ) 5.

The function is also shifted 1 unit up from the origin; therefore, 1 is added to g x x= +( ) 5 : g x x= + +( ) 5 1 .

Based on this information, we can conclude that the radical equation represented in the graph is g x x= + +( ) 5 1.

Lesson 4: Graphing Radical FunctionsU3-159

PRACTICE

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPSLesson 4: Graphing Radical Functions

For problems 1–5, create an equation in two variables for each graphed function.

1. y

x

8

6

4

2

–2

–2 2 4 6 8 10 12 140

2.

–14 –12 –10 –8 –6 –4 –2 2 4

4

2

–2

–4

y

x

0

3. 6

4

2

–2

–2 2 4 6 8 10 12 14

y

x

0

Practice 3.4.1: Creating Radical Equations in Two Variables

continued


PRACTICE


4. 6

4

2

–2

–2 2 4 6 8 10 12 140

y

x

5. 4

2

–2

–4

–10 –8 –6 –4 –2 2 4 6 8 100

y

x

For problems 6–9, create an equation in two variables for each function described.

6. A square root function is shifted 6 units to the left and 2 units up from the origin.

7. A cube root function is shifted 3 units down from the origin.

8. A square root function is shifted 4 units to the right of the origin and flipped over the x-axis.

9. A cube root function is shifted 2 units up and 3 units to the left of the origin.

continued


PRACTICE


Use your knowledge of radical equations to complete the problem that follows.

10. Explain why the following graph is not the graph of y x= + +4 8 .

2

–2

–4

–6

–8

–2 2 4 6 8 10 12 140

y

x


Introduction

A radical function can be graphed by identifying some of its key features and characteristics, such as its domain, x-intercept(s), and y-intercept(s). By knowing these features of a radical function, you can visualize what the graph will look like prior to graphing by hand or by using technology.

Key Concepts

• Recall that a radical function is a function that includes a radical sign as part of the equation.

• The parent function of the square root function family is f x x=( ) and the parent function of the cube root function family is f x x=( ) 3 . All other functions of the square root and cube root function families are based on these basic forms.

• The key features of a radical function are the characteristics that can be used to sketch the graph of the function without using a table of values.

• Key features include the x- and y-intercepts as well as the domain of the function.

• The x-intercept is the point at which a graphed function crosses the x-axis; it is written in the form (x, 0).

• The y-intercept is the point at which a graphed function crosses the y-axis, and is written in the form (0, y).

• The domain of a function is the set of all input values (x-values) that satisfy the given function without restriction.

• Recall that the domain of a function can be expressed in interval notation. That is, the domain is written in the form of (a, b), where a and b are the endpoints of the interval.

• Intervals can be written in different ways, depending on whether endpoints are included. For example, an interval of (2, 10) includes all numbers between 2 and 10, but not the endpoints. An interval of [2, 10] includes all numbers between 2 and 10, including the endpoints. And, an interval with mixed notation, such as (2, 10], includes all numbers between 2 and 10; 2 is not included, but 10 is included.

Lesson 3.4.2: Graphing Radical Functions


Identifying the Key Features of a Square Root Function

• The expression under a square root cannot result in a negative value; therefore, the domain of a square root function can be found by setting the expression under the radical to be greater than or equal to 0 and solving for x. This indicates the possible values of x for the graph of the function.

• The x-intercept can be found by setting the expression under the radical equal to 0 and solving for x.

• The y-intercept can be found by evaluating f(0). This determines the output, or y-value, for an input, or x-value, of 0.

Identifying the Key Features of a Cube Root Function

• The domain of a cube root function can be either positive or negative, meaning the domain is always all real numbers or (–∞, ∞); therefore, the values of x are not limited.

• Like with square root functions, the x-intercept can be found by setting the expression under the radical equal to 0 and solving for x.

• Also like with square root functions, the y-intercept can be found by evaluating f(0). This determines the output, or y-value, for an input, or x-value, of 0.

• Graphs of sketched functions can be verified by using a graphing calculator.

On a TI-83/84:


Step 2: At Y1, use your keypad to enter values for the function. Use [X, T, θ, n] for x and [x2] for any exponents. Press [GRAPH].

Step 3: To view a table of values for the function, press [2ND][GRAPH].

On a TI-Nspire:




Step 4: To see a table of values, press [menu], then use the arrow keys to select 2: View, then 9: Show Table. Press [enter].



Identify the key features for the graph of f x x=( ) .

1. Determine the domain of the function.

To find the domain of the function, set the expression underneath the radical sign to be greater than or equal to 0 and solve for x.

The expression under the radical sign is x; therefore, x ≥ 0.

The domain of the function is x ≥ 0. Because 0 is included in the domain, use a bracket for the lower bound. Since the domain continues to infinity, use a parenthesis for the upper bound: [0, ∞).

2. Determine the x-intercept of the function.

To find the x-intercept of the function, set the expression under the radical sign equal to 0 and solve for x.

The expression under the radical is x; therefore, x = 0.

The x-intercept is in the form (x, 0); therefore, the x-intercept of this function is (0, 0).

3. Determine the y-intercept of the function.

To find the y-intercept, evaluate f(0).

f x x=( ) Original function

f =(0) (0) Substitute 0 for x.

f(0) = 0

The y-intercept of this function is (0, 0).

4. Summarize the key features.

The domain is x ≥ 0, or [0, ∞). The x-intercept is (0, 0). The y-intercept is (0, 0).


Example 2

Identify the key features of the function g x x= +( ) 2 , then use them to create a graph.



The expression under the radical sign is x + 2; therefore, evaluate x + 2 ≥ 0.

x + 2 ≥ 0

x ≥ –2

The domain of the function is x ≥ –2, or [–2, ∞).


To find the x-intercept of the function, set the expression underneath the radical sign to be equal to 0 and solve for x.

The expression under the radical sign is x + 2; therefore, x = –2.

The x-intercept is in the form (x, 0); therefore, the x-intercept is (–2, 0).


To find the y-intercept, evaluate g(0).

g x x= +( ) 2 Original function

g = +(0) (0) 2 Substitute 0 for x.

g =(0) 2 Simplify.

The y-intercept of the function is ( )0, 2 .


4. Graph the function.

Use the key features to create a sketch of the function.

Begin by plotting the x-intercept at (–2, 0).

Next, plot the y-intercept, ( )0, 2 .

It is often helpful to choose several values for x and calculate the corresponding y-values to add to the shape of the graph.

Based on the domain, we know the function will continue increasing as x approaches positive infinity.

Draw a smooth curve through the graphed points, using an arrow at the end of the graphed curve opposite the x-intercept to indicate that the function continues toward infinity.

The function g x x= +( ) 2 is graphed as follows.

–2 2 4 6 8 10 12 14

6

4

2

–2

0

y

x


Example 3

Identify the key features of the function g x x=−( ) , then use them to create a graph.



The expression under the radical sign is x; therefore, x ≥ 0.

The domain of the function is x ≥ 0, or [0, ∞).


To find the x-intercept of the function, set the expression under the radical sign equal to 0 and solve for x.

The expression under the radical is x; therefore, x = 0.

The x-intercept is in the form (x, 0); therefore, the x-intercept of this function is (0, 0).


To find the y-intercept, evaluate g(0).

g x x=−( ) Original function

g =−(0) (0) Substitute 0 for x.

g =(0) 0 Simplify.

The y-intercept of the function is (0, 0).




Begin by plotting the x-intercept at (0, 0).

(0, 0) is also the y-intercept of the function.

Notice the negative sign in front of the radical. This indicates that the function will be similar to the parent function f x x=( ) , but will be reflected over the x-axis.


Choose several other values for x to calculate the corresponding y-values.

For example, let x = 1, 4, and 9. The corresponding y-values are –1, –2, and –3. These points, (1, –1), (4, –2), and (9, –3) can then be plotted.

Draw a smooth curve through the graphed points, using an arrow at the end opposite the x-intercept to indicate that the function continues toward infinity.

The function g x x=−( ) is graphed as follows.

6

4

2

–2

–4

–6

–2 2 4 6 8 10 12 140

y

x


Example 4

Identify the key features of the function f x x= +( ) 2 3, then use them to create a graph.



The expression under the radical sign is x + 3; therefore, evaluate x + 3 ≥ 0.

x + 3 ≥ 0

x ≥ –3

The domain of the function is x ≥ –3, or [–3, ∞).



The expression under the radical sign is x + 3; therefore, x = –3.

The x-intercept is in the form (x, 0); therefore, the x-intercept of this function is at (–3, 0).



f x x= +( ) 2 3 Original function

f = +(0) 2 (0) 3 Substitute 0 for x.

f =(0) 2 3 Simplify.

The y-intercept of the function is ( )0,2 3 , or approximately (0, 3.46).




Begin by plotting the x-intercept at (–3, 0).

Next, plot the y-intercept, (0, 3.46).


If necessary, choose additional values of x and calculate the corresponding y-values to obtain additional points.

Draw a smooth curve through the graphed points, using an arrow at the end opposite the x-intercept to indicate that the function continues toward infinity.

The function f x x= +( ) 2 3 is graphed as follows.

10

8

6

4

2

–2

–4 –2 2 4 6 8 10 12 140

y

x


Example 5

Identify the key features of the function f x x= −( ) 43 , then use them to create a graph.


Since cube roots can be positive or negative, there are no limitations on x; therefore, the domain of x is all real numbers or (–∞, ∞).



The expression under the radical sign is x – 4; therefore, x = 4.

The x-intercept is in the form (x, 0); therefore, the x-intercept of this function is at (4, 0).



f x x= −( ) 43 Original function

f = −(0) (0) 43 Substitute 0 for x.

f = −(0) 43 Simplify.

f(0) ≈ –1.59

The y-intercept of the function is (0, –1.59).




Begin by plotting the x-intercept at (4, 0).

Next, plot the y-intercept, (0, –1.59).

Based on the domain, (–∞, ∞), we know the graph will continue toward positive and negative infinity.

Choose additional values for x to calculate the corresponding y-values.

For example, let x = –2, 2, and 6. The corresponding y-values are approximately –1.87, –1.26, and 1.59. These points, (–2, –1.87), (2, –1.26), and (6, 1.59) can then be plotted.

The function f x x= −( ) 43 is graphed as follows.

10

8

6

4

2

–2

–4

– 4 –2 2 4 6 8 10 12 14

y

x

0


PRACTICE


For problems 1–3, identify the domain, x-intercept, and y-intercept for each function. Round answers to the nearest tenth.

1. t x x=− +( ) 2 1

2. f x x= −( ) 2

3. g x x= −( ) 53

For problems 4–7, identify the domain, x-intercept, and y-intercept for each function. Round answers to the nearest tenth. Then, use the information to graph each function.

4. h x x=−( ) 2 3

5. f x x=− + −( ) 3 1

6. f x x= −( ) 3 6

7. g x x=( ) 4

For problems 8–10, identify the domain for each function. Then, use the information to complete each problem.

8. What is the domain of the function representing the speed of sound, S t= +21.9 5 2457 ?

9. Consider the functions f x x=( ) and g x x=( ) 3 . Which function has a domain of all real numbers?

10. How do the graphs of f x x=( ) and g x x=−( ) compare in terms of their domains?

Practice 3.4.2: Graphing Radical Functions

Lesson 5: Comparing Properties of Functions



Common Core Georgia Performance Standard

MCC9–12.F.IF.9

Essential Questions

1. How can you find the vertical and horizontal asymptotes of a rational function?

2. How can you find the domains and ranges of rational and radical functions?

3. How can you find the zero(s) and y-intercept(s) of a function?

4. How can you compare functions in different formats, such as tables, graphs, and equations?

WORDS TO KNOW

asymptote a line that a function gets closer and closer to, but never crosses or touches

domain the set of all input values (x-values) that satisfy the given function without restriction

range the set of all outputs of a function; the set of y-values that are valid for the function

root the x-intercept of a function; also known as zero

x-intercept the point(s) where a function equals 0 and at which the graph crosses the x-axis; also called root or zero

y-intercept the point at which the graph crosses the y-axis; written as (0, y)

zero the x-intercept of a function; also known as root

U3-175Lesson 5: Comparing Properties of Functions


• Khan Academy. “Asymptotes of Rational Functions.”


Site visitors can watch any or all of the three provided videos that explain how to find vertical and horizontal asymptotes and include examples. Horizontal asymptotes of higher degree polynomials are also discussed.

• Khan Academy. “Domain of a Radical Function.”


Visitors of this site can watch a video that explains how to find the domain of a radical function. Other companion videos give additional instruction on how to find the domain and range of a function.

• Purplemath.com. “Graphing Overview: Polynomials, Radicals, Rationals, and Piecewise.”


This site provides a general overview of the graphs of various functions, including rational and radical functions, as well as links to additional information specific to each type of function.

• Virtual Math Lab, West Texas A&M University. “Tutorial 40: Graphs of Rational Functions.”


In this enrichment activity, a written tutorial discusses how to find the asymptotes of a rational function and then graph the function. Higher-degree polynomials are discussed, and users are taught how to find oblique asymptotes as well.


Lesson 3.5.1: Comparing Properties of FunctionsIntroduction

Functions have many characteristics, such as domain, range, asymptotes, zeros, and intercepts. These functions can be compared even when given in a different format, such as when provided as an equation, graph, or table. The key to comparing functions is to either find the same information from the different forms or transfer the functions into the same form for a clear comparison.

Key Concepts

• Functions can be described numerically in a table, verbally, algebraically, or graphically.

• To compare two functions, determine the possible zeros, y-intercepts, domain, and range of each function.

• Depending on how the function is represented, this information can be found in various ways.

Determining the Zeros of a Function

• The zero of a function is the point where a function equals 0 and at which the graph crosses the x-axis. A zero is also called the root or x-intercept of a function.

• To identify the zero of an equation, replace f(x) or y with 0 and solve for x.

• To identify the zero of a graph, look for the point where the curve crosses the x-axis.

• To identify the zero in a table, look for a point that has 0 as its y-coordinate.

Determining the y-intercepts of a Function

• The y-intercept of a function is the point where the graph crosses the y-axis. It is written as (0, y).

• To identify the y-intercept of an equation, replace x with 0 and solve for f(x) or y.

• To identify the y-intercept of a graph, look for the point where the curve or line crosses the y-axis.

• To identify the y-intercept in a table, look for a point that has 0 as its x-coordinate.


Determining the Asymptotes of a Function

• An asymptote is a line that a function gets closer and closer to, but never crosses or touches. In other words, it is a line that a curve approaches (but does not reach) as its x- or y-values become very large or very small.

• When a rational function is in equation form, find the vertical asymptote by setting the denominator equal to 0 and solving for x.

• To identify the vertical asymptote of a graph, look for a vertical line that the curve approaches but never reaches.

• To identify the vertical asymptote in a table, look for a grouping of x-coordinates with similar values.

• For a first-degree rational function, the horizontal asymptote can be found

by dividing the coefficients of the variables. For example, in the function

f xx

x( )

3 1

2 5=

−+

, 3 divided by 2 is 1.5, and the asymptote is located at y = 1.5.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –10

1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

f(x)Asymptote y = 1.5

• To identify the horizontal asymptote of a graph, look for a horizontal line that the curve approaches but never reaches.

• To identify the horizontal asymptote in a table, look for a grouping of y-coordinates with similar values.


Determining the Domain of a Function

• The domain of a function is the set of all input values (x-values) that satisfy the given function without restriction.

• For a radical function, the domain of an equation is found by solving an inequality in which the radicand is listed as being greater than or equal to 0.

• The domain of a first-degree rational function is found by identifying the asymptotes. The domain will contain every real number except the point where the vertical asymptote is defined. To identify the domain from a graph, determine the x-coordinate where the curve starts and in which direction the curve continues. Whichever values are not allowed in the domain will be vertical asymptotes on the graph.

• Recall that the domain of a function can be expressed in interval notation. That is, the domain is written in the form of (a, b), where a and b are the endpoints of the interval. Depending on the values of the interval, the notation may change, as shown in the following table.

Interval notation

Example Description

(a, b) (2, 10)All numbers between 2 and 10; endpoints are not included.

[a, b] [2, 10]All numbers between 2 and 10; endpoints are included.

(a, b] (2, 10]All numbers between 2 and 10; 2 is not included, but 10 is included.

[a, b) [2, 10)All numbers between 2 and 10; 2 is included, but 10 is not included.

Determining the Range of a Function

• The range is the set of all outputs of a function. It is the set of y-values that are valid for the function.

• For a radical function, the best way to find the range of an equation is to first identify the domain, and then find the corresponding values for f(x) or y.

• The range of a first-degree rational function is found by identifying the asymptotes. The range will contain every real number except the point where the horizontal asymptote is defined.

• To identify the range from a graph, determine the y-coordinate where the curve starts and in which direction the curve continues.

• Similar to the domain, the range can also be expressed in interval notation.


Example 1

Given f(x) as shown in the graph and g xx

x( )

3 6

7 2=

−+

, which function has the greater zero?

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

y

–12 –10 –8 –6 –4 –2 2 4 6 8 10 12

x

f(x)

0

1. Identify the zero of f(x).

Find the zero of a graph by identifying the point at which the curve crosses the x-axis.

The curve of the graph crosses the x-axis at x = 5.

Therefore, 5 is the zero of f(x).



2. Identify the zero of g(x).

To find the zero of the function g(x), set the function equal to 0 and solve for x.

g xx

x( )

3 6

7 2=

−+

Original function

x

x(0)

3 6

7 2=

−+

Substitute 0 for g(x).

0(7x + 2) = 3x – 6 Cross multiply.

0 = 3x – 6 Simplify.

6 = 3x Add 6 to both sides.

x = 2 Divide both sides by 3.

The zero of g(x) is 2.

3. Determine which function has the greater zero.

The zero of f(x) is 5 and the zero of g(x) is 2.

5 is greater than 2; therefore, given the two functions f(x) and g(x), f(x) has the greater zero.


Example 2

Grady must graph both f(x) and g(x) over the interval [–20, 20]. He has graphed f(x) as shown below, and now he must graph g x x( ) 2 3 4= + + . Which of these two functions has the larger domain?

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

y

–2 2 4 6 8 10 12 14 16 18 20

x

f(x)

0

1. Identify the domain of f(x).

To find the domain of a graph, identify the x-coordinates where the curve starts and stops.

The curve starts at the point (4, –1) and continues towards infinity.

The x-coordinate is 4, so the domain appears to be [4, )∞ . However, since the graph is only over the interval [–20, 20], the domain stops at 20.

Thus, the domain of f(x) is [4, 20].


2. Identify the domain of g(x).

g(x) is presented as an equation where the radicand must be greater than or equal to 0.

To find the domain of g(x), create an inequality in which the radicand is greater than or equal to 0 and solve for x.

The given function is g x x( ) 2 3 4= + + , where 3x + 4 is the radicand.

3x + 4 ≥ 0 Set the radicand greater than or equal to 0.

3x ≥ –4 Subtract 4 from both sides.

x4

3≥−

Divide both sides by 3.

x 11

3≥− Convert the improper fraction to a mixed fraction.

The domain of g(x) would be 11

3,− ∞

, but since the graph is only

over the interval [–20, 20], the domain of g(x) is 11

3, 20−

.


3. Determine which function has the larger domain.

First determine how wide each interval is.

The domain of f(x) is [4, 20].

To determine how wide the interval is, find the difference of the endpoints.

For this interval, the endpoints are 20 and 4.

20 – 4 = 16

The interval [4, 20] is 16 units wide.

The domain of g(x) is 11

3, 20−

.

To determine how wide this interval is, find the difference of the endpoints.

For this interval, the endpoints are 20 and 11

3− .

20 11

321

1

3− −

=

The interval 11

3, 20−

is 21

1

3 units wide.

f(x) is 16 units wide.

g(x) is 211

3 units wide.

211

3 is greater than 16; therefore, g(x) has the larger domain.


Example 3

The curve of the rational function f(x) passes through the origin and approaches x = 3 and y = –2 as it approaches infinity. Which asymptotes, if any, does f(x) share with the

function g xx

x( )

6 10

3 9=− −

+?

1. Identify the asymptotes of f(x).

Since the problem statement says that the curve approaches x = 3 and y = –2 as it approaches infinity, this means x = 3 and y = –2 are the asymptotes.

Since x = 3 is a vertical line, it is the vertical asymptote of f(x).

Since y = –2 is a horizontal line, it is the horizontal asymptote of f(x).

2. Identify the vertical asymptote of g(x).

g(x) is a rational function.

To find the vertical asymptote of a rational function, set the denominator equal to 0 and solve for x.

The denominator of the function g xx

x( )

6 10

3 9=− −

+ is 3x + 9.

3x + 9 = 0 Set the denominator of g(x) equal to 0.

3x = –9 Subtract 9 from both sides.

x = –3 Divide both sides by 3.

The vertical asymptote of g(x) is located at x = –3.


3. Identify the horizontal asymptote of g(x).

g(x) is a first-degree rational function.

To find the vertical asymptote of a first-degree rational function, divide the coefficients of the variables.

In the function g xx

x( )

6 10

3 9=− −

+, the terms with variables are –6x

and 3x.

The coefficient of the variable in the numerator is –6.

The coefficient of the variable in the denominator is 3.

y6

32=

−=−

The horizontal asymptote of g(x) is located at y = –2.

4. Determine which asymptotes the functions share, if any.

The vertical asymptote of f(x) is located at x = 3, and the vertical asymptote of g(x) is located at x = –3. These functions do not share not the same asymptote, since one is positive and one is negative.

The horizontal asymptote of f(x) is located at y = –2, and the horizontal asymptote of g(x) is located at y = –2. This is the same asymptote.

Both functions have the same horizontal asymptote at y = –2.


Example 4

The radical function f(x) has a domain of [ 4, )− ∞ and a range of [5, )∞ . The function g(x) is shown in the following graph. Which function has a higher y-intercept?

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

y

–16 –14 –12 –10 –8 –6 –4 –2 2 4 6

x0

g(x)

1. Estimate the y-intercept of g(x).

The curve of the graph appears to cross the y-axis at the point (0, 2.5).

2. Estimate the y-intercept of f(x).

Since the range of the function is [5, )∞ , the function only exists at or above y = 5. Therefore, this function must intercept the y-axis somewhere above the point (0, 5).

3. Determine which function has a higher y-intercept.

Even though the exact y-intercept of f(x) cannot be determined, it is clear that it is located somewhere above the point (0, 5). Since the point (0, 2.5) is the y-intercept of g(x), and 2.5 is less than 5, it can be determined that f(x) has the higher y-intercept.


Example 5

What are the similarities and differences between the ranges of f xx

x( )

4 5=− +

and

g(x), a first-degree rational function that contains the following coordinates?

x –6 –5 –4 –3 –2 0 1 2 3 4 5g(x) 2.4 2.5 2.7 3 4 0 1 1.3 1.5 1.6 1.7

1. Identify the range of f(x).

The range for a first-degree rational function is every real number except for the location of the horizontal asymptote.

The horizontal asymptote is found by dividing the coefficients of the variables.

In the function f xx

x( )

4 5=− +

, the terms with variables are –4x and x.

The coefficient of the variable in the numerator is –4.

The coefficient of the variable in the denominator is an understood 1.

y4

14=

−=−

The horizontal asymptote of f(x) is located at y = –4.

Therefore, the range of f(x) is all real numbers except y = –4.

2. Identify the range of g(x).

The range for a first-degree rational function is every real number except for the location of the horizontal asymptote.

The values for g(x), or y, in the table approach 2 on both ends, so a reasonable estimate for the horizontal asymptote is y = 2.

Therefore, the range of g(x) is all real numbers except y = 2.

3. Explain the similarities and differences between the ranges.

For both functions, the range contains all real numbers except for one. However, the functions have different exceptions. For f(x), the exception is y = –4; for g(x), the exception has been approximated as y = 2.


PRACTICE

UNIT 3 • RATIONAL AND RADICAL RELATIONSHIPSLesson 5: Comparing Properties of Functions

For problems 1–3, f(x) is defined as f xx

x( )

3

4=

− and g(x) is a rational function, with

some values shown in the table below.

x g(x)–10 2.67–8 2.57–4 2–2 0

–1.3 –7–0.25 7

0 62 45 3.5

10 3.27

1. Which function, f(x) or g(x), has the higher y-intercept? Explain your reasoning.

2. Which function, f(x) or g(x), has the greater zero? Explain your reasoning.

3. Do the functions share any vertical or horizontal asymptotes? If so, name the asymptote(s).

Practice 3.5.1: Comparing Properties of Functions

continued

Lesson 5: Comparing Properties of FunctionsU3-189

PRACTICE


For problems 4–6, the graph of f(x) is shown below and g(x) is defined as g x x( ) 4 1=− − + .

6

8

4

2

–2

–4

–6

–8

y

–6 –4 –2 2 4 6 8 10 12

x

0

f(x)

4. Which function, f(x) or g(x), has the higher y-intercept? Explain your reasoning.

5. Which function, f(x) or g(x), has the greater zero? Explain your reasoning.

6. What domain do both functions share? Explain your reasoning.

continued


PRACTICE


For problems 7–10, f xx

x( )

2 7

2 4=− +

−, and g(x) is a rational function that goes through

the origin and whose curve approaches infinity as it approaches x = –5 and y = 3.

7. How far apart are the vertical asymptotes of the two functions? Explain your reasoning.

8. Which function has a higher horizontal asymptote? Explain your reasoning.

9. Which function has a greater zero? Explain your reasoning.

10. What is the difference between the ranges of the functions?

AK-1Answer Key

Answer KeyLesson 1: Operating with Rational Expressions

Practice 3.1.1: Adding and Subtracting Rational Expressions, pp. U3-15–U3-16

1. x

7

2.

3. x

xx

2 50

25, 5

2

2

+−

≠ ±

4.

5. x x x

x

2 5 1

1

3 2

2

− + −+

6.

7. x x

x xx x

9 8 38

6 19 10,

5

2,

2

3

2

2

+ −− +

≠ ≠

8.

9. 7 7 7 2

2centimeters, 1

3 2

2

x x x

x xx

+ + ++

>

Practice 3.1.2: Multiplying Rational Expressions, pp. U3-23–U3-24

1. x x

xx

2 9 10

2, 0

2 + +≠

2.

3. x

x x x5

1, 8,

3

2, 1

−≠ − ≠ − ≠

4.

5. x

xx x x

2 3

1, 4, 3, 12

−+

≠ − ≠ − ≠ −6.

7. 2

9 3

3 6

4

( 2)

3 (3 )

3 ( 2)

( 2) ( 2) 3 3,

2

2

x x

x

x

x

x x

x

x

x x

x

x

x

x

−+

•+−

=−+

•+

+ −=

+=

+ x ≠ –3

8.

9. x x

x xx

5 10 5

3centimeters , 3

2

3 22

+ +−

>

Practice 3.1.3: Dividing Rational Expressions, pp. U3-35–U3-36

1. x

xx

6 21

5, 0

−≠

2.

3. x x

xx x

5

2 6, 3, 0

2 −+

≠ − ≠

4.

5. x x

x x

x

x

x x

x x

x

x

x x

x x

x

x

15 27 6

2

5 1

3 3

15 27 6

2•

3 3

5 1

3 ( 2) (5 1)

( 2) ( 1)•

3 ( 1)

(5 1)9

2

2

2

2

+ −+ −

÷−−

=+ −+ −

−−

=+ −+ −

−−

=

6.

7. xx

x( 4)11

1, 1+ +

−≠

8.

9. x

x x

9

2

2

2

−−

grams per meter3, x > 3

AK-2Answer Key

Lesson 2: Solving Rational and Radical Equations

Practice 3.2.1: Creating and Solving Rational Equations, pp. U3-51–U3-521. x = 202.

3. x = –44.

5. t t

+ =120 90

1

6.

7. t t+ =

120

2

160

4600

8.

9. x = 39

13

Practice 3.2.2: Creating and Solving Rational Inequalities, pp. U3-69–U3-701. 5 < x2.

3. –6 < x < –1 or x > 24.

5. One inequality that works is t t

t

+ −<

4 25 2.730

2

.6.

7. They can rake the yards for all the houses on the cul-de-sac and still have time for more if they spend at least 7.2 hours working.8.

9. The concentration is less than 0.01 mg/mL if 0 ≤ t ≤ 2 or if t > 4; because the concentration rises to unsafe levels after 2 days but before 4 days, at least 4 days must pass before safety signs can be removed.

Practice 3.2.3: Solving Radical Equations, pp. U3-81–U3-831. x = –62.

3. x = –54.

5. v = 2.4 m/s6.

7. g = 9.8 m/s28.

9. The pendulum length is about 20 m.


Practice 3.3.1: Creating Rational Equations in Two Variables, pp. U3-97–U3-98

1. Answers may vary. Sample answer: ( )3

3=−+

f xx

x or ( )

3

3=

+−

f xx

x2.

3. Answers may vary. Sample answer: ( )3

3 22=−−

f xx

x or ( )

3

2 3 2=−

f xx

x4.

5. ( )3 22

=−

f nn

n6.

7. The average rate is half of the sum of the rates for each leg of the trip: 1

2( )

1

2

100 100

2

100 100

21 2 2+ = ++

=

++

r rt t

t

t t.

8.

9. Answers may vary. Sample answer: ( )( 4)( 8)

2=

− −+

f xx x

x

AK-3Answer Key

Practice 3.3.2: Graphing Rational Functions, pp. U3-112–U3-1141.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

2.

3.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

–10

–9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

4.

5. asymptotes: x = –3, y = x – 1; no real zeros6.

7. ( )2 23π

=+

S rr

r8.

9. As the radius r approaches 0.4, the surface area approaches a minimum value.

AK-4Answer Key

Practice 3.3.3: Finding the Zeros, p. U3-1291. There is one real zero: x = –1.

2.

3. zeros: 5 and7

2= = −x x

4.

5. zero: x = –36.

7. no real zeros8.

9. It does not apply to the real-world situation because it is negative.

Practice 3.3.4: Solving a System of Rational Equations, pp. U3-144–U3-145

1. 3,4

9, (1, 0)−

2.

3. infinitely many solutions4.

5. 2

2, 1 ,

2

2, 1−

− −

6.

7. infinitely many solutions8.

9. At a value of 26

2= +t hours, the time needed by 2 goats is the same as that needed by 3 goats.

Lesson 4: Graphing Radical Functions

Practice 3.4.1: Creating Radical Equations in Two Variables, pp. U3-159–U3-161

1. y x= − +3 12.

3. y x= −14.

5. y x= +336.

7. y x= −33 8.

9. y x= + +3 23

Practice 3.4.2: Graphing Radical Functions, p. U3-1731. domain: [–0.5, ∞); x-intercept: (–0.5, 0); y-intercept: (–1, 0)2.

3. domain: (–∞, ∞); x-intercept: (5, 0); y-intercept: (0, –1.71)4.

5. domain: [1, ∞); x-intercept: (10, 0); y-intercept: none

–4 –2 2 4 6 8 10 12 14

10

8

6

4

2

–2

–4

–6

–8

–10

0

y

x

6.

AK-5Answer Key

7. domain: [0, ∞); x-intercept: (0, 0); y-intercept: (0, 0)

–6 –4 –2 2 4 6

10

8

6

4

2

–2

–4

y

x

0

8.

9. g(x) has a domain of all real numbers.

Lesson 5: Comparing Properties of Functions

Practice 3.5.1: Comparing Properties of Functions, pp. U3-188–U3-1901. g(x) has a higher y-intercept; the y-intercept of f(x) is at (0, 0) and the y-intercept of g(x) is at (0, 6).2.

3. f(x) and g(x) share a horizontal asymptote at y = 3.4.

5. g(x) has a greater zero; the zero of f(x) is between –1 and –2 and the zero of g(x) is at (1, 0).6.

7. 7 units; the vertical asymptote of f(x) is at x = 2 and the vertical asymptote of g(x) is at x = –5. The difference between –5 and 2 is 7 units.8.

9. f(x) has a greater zero; the zero of f(x) is located at the point (3.5, 0) and the zero of g(x) is most likely less than –5, since the graph approaches infinity at the point (0, –5).

mathplayer.weebly.commathplayer.weebly.com/uploads/3/8/0/0/38006537/ccgps_advanced_alg… · v i...

Documents

Transcript of mathplayer.weebly.commathplayer.weebly.com/uploads/3/8/0/0/38006537/ccgps_advanced_alg… · v i...